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Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution Logarithmic equation Page number 92 (Q. No. 125 to 261) 125. log x – 1 3 = 2 x – 1 1 and x – 1 > 0 (x – 1) 2 = 3 x 2 x > 1 x 2 + 1 – 2x = 3 x 2 – 2x – 2 = 0 D = b 2 – 4ac = 12 Roots = a 2 D b + = 2 3 2 2 ± = 2 ) 3 1 ( 2 ± = 1 ± 3 Since 3 1+ satisfies the equation Final solution x { 3 1+ } 126. log 4 (2 log 3 (1 + log 2 (1 + 3 log 3 x))) = 2 1 2 1 log 2 (2 log 3 (1 + log 2 (1 + 3 log 3 x))) = 2 1 2 log 3 (1 + log 2 (1 + 3 log 3 x))) = 2 1 = 2 (1 + log 2 (1 + 3 log 3 4) = 3 1 log 2 (1 + 3 log 3 x) = 3 – 1 1 + 3 log 3 x = 2 2 3 log 3 x = 3 log 3 x = 1 x = 3 1 = 3 Final solution x {3} 127. log 3 (1 + log 3 (2 x – 7) = 1 1 + log 3 (2 x – 7) = 3 (2 x – 7) = 3 2 2 x = 9 + 7 = 16 x = 4 Final solution x {4} 128. log 3 (3 x – 8) = 2 – x 3 2 – x = 3 x – 8 x 2 3 3 = 3 x – 8 x 3 9 = 3 x – 8 Let 3 x = t t 9 = t – 8 9 = t 2 – 8t t 2 – 8t – 9 = 0 D = b 2 – 4ac = 64 + 36 = 100 t = 3 x = 2 10 8 ± = 2 18 , – 2 2 = 9 ,–1 (negative value is not possible as a x >0 ) 3 x = 9 3 x = 3 2 x = 2 Final solution x {2} 129. x 3 ) 2 9 ( log x 2 = 1 x 3 log 2 (9 – 2 x ) = 3 – x 9 – 2 x = (2) 3 – x 9 – 2 x = x 3 2 2 t 8 = 9 – t Now, 2 x = 1, 8 8 = 9t – t 2 t 2 – 9t + 8 = 0 (t – 1) (t – 8) = 0 t = 1, 8 x = 0, 3 but x 3 Final solution x {0} 130. log 5 – x (x 2 – 2x + 65) = 2 (5 – x) 2 = x 2 – 2x + 65 25 – x 2 – 10x = x 2 – 2x + 65 25 – 65 = 8x – 40 = 8x x = – 5 Final solution x {– 5} 131. log 3 (log 9 x + 2 1 + 9 x ) = 2x log 3 (log 9 x + 2 1 + 3 2x ) = 2x 2 1 = 2 1 log 3 x –1 = log 3 x x = 3 1 Final solution x { 3 1 }

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Page 1: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Logarithmic equation Page number 92 (Q. No. 125 to 261)

125. logx – 1 3 = 2 x – 1 ≠ 1 and x – 1 > 0⇒ (x – 1)2 = 3 x ≠ 2 x > 1⇒ x2 + 1 – 2x = 3⇒ x2 – 2x – 2 = 0

D = b2 – 4ac = 12

Roots = a2Db +−

= 2

322 ± =

2)31(2 ±

= 1 ± 3

Since 31+ satisfies the equation ∴∴∴∴∴ Final solution x ∈{

31+

}

126. log4(2 log3(1 + log2(1 + 3 log3x))) = 21 ⇒

21

log2(2 log3 (1 + log2 (1 + 3 log3x))) = 21

⇒ 2 log3 (1 + log2(1 + 3 log3x))) = 21 = 2 ⇒ (1 + log2 (1 + 3 log34) = 31

⇒ log2 (1 + 3 log3x) = 3 – 1 ⇒ 1 + 3 log3x = 22

⇒ 3 log3x = 3 ⇒ log3x = 1⇒ x = 31 = 3 ∴∴∴∴∴ Final solution x ∈{3}

127. log3 (1 + log3 (2x – 7) = 1 ⇒ 1 + log3 (2x – 7) = 3 ⇒ (2x – 7) = 32

⇒ 2x = 9 + 7 = 16 ⇒ x = 4 ∴∴∴∴∴ Final solution x

{4}

128. log3 (3x – 8) = 2 – x ⇒ 32 – x = 3x – 8 ⇒

x

2

33

= 3x – 8 ⇒ x39

= 3x – 8

Let 3x = t

t9

= t – 8 ⇒ 9 = t2 – 8t ⇒ t2 – 8t – 9 = 0

D = b2 – 4ac = 64 + 36 = 100

t = 3x = 2108 ±

= 2

18, –

22

= 9 ,–1 (negative value is not possible as ax >0 )

3x = 9 ⇒ 3x = 32 ⇒ x = 2 ∴∴∴∴∴ Final solution x ∈{2}

129.

x3)29(log x

2

−−

= 1 x ≠ 3

⇒ log2 (9 – 2x) = 3 – x ⇒ 9 – 2x = (2)3 – x ⇒ 9 – 2x = x

3

22 ⇒ t

8 = 9 – t

Now, 2x = 1, 8 ⇒ 8 = 9t – t2 ⇒ t2 – 9t + 8 = 0⇒ (t – 1) (t – 8) = 0 ⇒ t = 1, 8 ⇒ x = 0, 3 but x ≠ 3

∴∴∴∴∴ Final solution x ∈{0}

130. log5 – x (x2 – 2x + 65) = 2 ⇒ (5 – x)2 = x2 – 2x + 65 ⇒ 25 – x2 – 10x = x2 – 2x + 65

⇒ 25 – 65 = 8x ⇒ – 40 = 8x ⇒ x = – 5∴∴∴∴∴ Final solution x

{– 5}

131. log3 (log9x +

21

+ 9x) = 2x ⇒ log3 (log9x + 21

+ 32x) = 2x ⇒ – 21

= 21

log3x

–1 = log3x ⇒ x = 31

∴∴∴∴∴ Final solution x ∈{

31

}

Page 2: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

132. log3(x + 1) (x + 3) = 1 ⇒ 31 = (x + 1) (x + 3) ⇒ 3 = x2 + x + 3x + 33 = x2 + 4x + 3 ⇒ 0 = x2 + 4x ⇒ 0 = x(x + 4)∴ x = –4, 0 since – 4 doesn’t satisfy equation ∴∴∴∴∴ Final solution x ∈{0}

133. log7 (2x – 1) + log7 (2

x – 7) = 1Let 2x = tlog7(t – 1) (t – 7) = 1 ⇒ 71 = t2 – 7t – t + 7 ⇒ 0 = t2 – 8t ⇒ t = 0, 8

2x = 8 ⇒ x = 32x = 0 Impossible ∴∴∴∴∴ Final solution x

{3}

134. log 5 + log(x + 10) – 1 = log (2/x – 20) – log (2x – 1) ⇒ log (5 + x + 10) – log1010 = log

1x220x21

−−

⇒ log 10)10x(5 +

= log 1x220x21

−−

⇒210x +

= 1x220x21

−−

⇒ (2x – 1) (x + 10) = 42x – 40 ⇒ 2x2 – x + 20x – 10 = 42x – 40⇒ 2x2 – 19x – 42x – 10 + 40 = 0 ⇒ 2x2 – 20x – 3x + 30 = 0

⇒ (2x – 3) (x – 10) = 0 ⇒ x = 23

, 10

∴∴∴∴∴ Final solution x ∈{

23

, 10 }

135. 1 – log 5 = 31

++ 5log31xlog

21log ⇒ log10 10 – log105 = 3

1 [log102–1 + log10x + log1051/3]

⇒ log10 510

= 31

[log102–1 + log10x + log1051/3] ⇒ log1023 = 31

log102–1 + log10x + log1051/3

⇒ log1023 = log10

3 5x21

⇒ 8 = 2

5x 3⇒ x = 3 5

16

∴∴∴∴∴ Final solution x ∈{

3 516

}

136. log x – 21

log

−21x = log

+21x –

21

log

+81x ⇒ log x – log

21x − = log x +

21

– log 81x +

⇒ log

21x

x

− = log

81x

21x

+

+⇒

2

21x

x

− =

2

81x

21x

+

+⇒

21x

x2

− =

81x

x41x2

+

++

⇒ x3 + 8x2

= x3 + 4x

+ x2 – 2x2

– 81

– 2x

⇒ 8(8 x3 + x2) = 8(8x3 + 2x + 8x2 – 4x2 – 1 – 4x)

⇒ 8x3 + x2 = 8x3 – 2x + 4x2 – 1 ⇒ 0 = 3x2 – 2x – 1 ⇒ 0 = 3x2 – 3x + x – 1

⇒ 0 = 3x (x –1) +1(x – 1) ⇒ x = – 31

, 1 ∴∴∴∴∴ Final solution x ∈{ –

31

}

137. xloglog33 – logx + log2x – 3 = 0 ⇒ xlog – log x + log2x – 3 = 0

⇒ log10 x – log10x + log102x – 3 = 0 ⇒21

log10x – log10x + (log10x)2 – 3 = 0

⇒ log10x – 2 log10x + 2(log10x)2 – 62 = 0 ⇒ 2(log10x)2 – log10x – 6 = 0

⇒ 2(log10x)2 – 4 log10x + 3 log10x – 6 = 0 ⇒ (2 log10x + 3) (log10x – 2) = 0

⇒ log10x = 2 or log10x = – 23

⇒ x = 102 = 100 or x = 10–3/2

Page 3: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

138. 12)2log()2(log 52

)2( −−+−− xxx = 102 log (x – 2) (Let log (x-2) = t )

⇒ log )125)(2xlog()2x(log2)2x( −−+−− = log 102log (x – 2)

⇒ log (x – 2) (log2(x – 2)+ log (x – 2)5 – 12)) = log10102 log(x – 2)

⇒ t [t2 + 5t – 12)] = 2t ⇒ t3 + 5t2 – 14t = 0 ⇒ t(t2 + 5t – 14) = 0⇒ t(t + 7)(t – 2) = 0 ⇒ t = 0, –7, 2 ⇒ log (x-2) = 0, –7, 2⇒ x = 2 ,2 + 10–7 , 102

139. )x21(log39 − = 5x2 – 5

⇒ )x21(log2 33 − = 5x2 – 5

⇒ 23 )x21(log3 − = 5x2 – 5

⇒ (1 – 2x)2 = 5x2 – 5⇒ 1 + 4x2 – 4x = 5x2 – 5⇒ 0 = 5x2 – 4x2 – 4x – 5 – 1⇒ 0 = x2 – 4x – 6

D = b2 – 4ac

= 16 + 24 = D40 = 102

α = a2

Db +− = 2 + 10 , 2 – 10

140. x1 + log x = 10xlogxx

1 + logx = log 10x

⇒ (log10x) (1 + log10x) = log1010 + log10x⇒ log10x + log10

2x = 1 + log10x⇒ log10

2x = 1⇒ log10 x = ±1

∴ x = 10–1, 101

141. x2 logx = 10x2

⇒ log10x2.log x = log1010 + log10x2

⇒ 2 log10x (log10x) = 1 + 2 log10x⇒ 2 log10

2x – 2 log10x – 1 = 0 Let log10x = t⇒ 2t2 – 2t – 1 = 0

2t2D = b2 – 4ac

= 4 + 8 = 12 D = 32

t = 9

322 ± = 4

)31(2 ± = 2

31±

Now, log10x = t = 2

31+,

231−

∴ x = 2

31±

⇒ log10x = 2

31±

⇒ log10x2 = 31±

Page 4: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

⇒ x2 = 3110 ±

x = 3110 + , 3110 −

142. 35xlog

x+

= 105 + log x

log x . 3

5xlog + = log10105 + log10x

⇒ log x . 35xlog + = 5 + log10x

⇒ log2x + 5 log10x = 15 + 3 log10x⇒ log10

2x + 2 log10x – 15 = 0⇒ log10x (log10x + 5) – 3(log10x + 5) = 0⇒ (log10x – 3) (log10x + 5) = 0⇒ log10x = 3 log10x = – 5

⇒ x = 103 x = 10–5

143. x.x 3log = 9

⇒ x.3log3 xlog = log39

⇒ log3x (log3x) = 21⇒ log3

2x = 21

⇒ log3x = ± 2

⇒ x = 23 , 23−

144. 1xlog5)x( − = 5

⇒ log5 x . log5 (x – 1) = log55

⇒21

log5x (log5x – 1) = 1

⇒21

log52x –

21

log5x = 1

⇒ = log52 x – log5x = 2

⇒ = log52 x – log5x – 2 = 0⇒ log5x (log5x + 1) – 2(log5x + 1) = 0⇒ (log5x – 2) (log5x + 1) = 0⇒ log5x = 2 and log5x = –1

⇒ x = 52 and x = 51

⇒ x = 5 and x = 51

145. xlogx +1 = 106

⇒ log x (log x + 1) = log10106

⇒ log102x + log10x – 6 = 0

⇒ (log10x – 2) (log10x + 3) = 0⇒ log10x = 2 log10x = – 3

Page 5: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution⇒ x = 102 x = 10–3

146. 47xlog

x+

= 10log x + 1

⇒ log x. log 4

7x + = 4 log10 x + 4

⇒ xlog210 + 3 log10x – 4 = 0

⇒ xlog210 + 4 log10x – log10x – 4 = 0

⇒ (log10x + 4) (log10x – 1) = 0⇒ x = 104 and x = 101

147.)2x(

xlogx−

= 9

⇒ )2x(log2/1 xx − = 9

⇒ 2x )2x(logx − = 9

⇒x

x xlog2)2x( − = 9

⇒⇒ (x – 2)2 – 9 = 0⇒ x2 + 4 – 4x – 9 = 0⇒ x2 – 4x – 5 = 0⇒ x(x – 5) + 1(x – 5) = 0⇒ x = –1, 5 but x > 0⇒ ∴ x= 5 is a solution

148.2xlogxlog 22

2xlog −+

= log x

⇒21 2xlogxlog 22

xlog −+ = 21

2 log x

⇒2xlogxlog 22

2xlog −+

= log x

⇒2xlogxlog 22

2xlog −+

= 2 log x

149. xlog3 2 – log28x + 1 = 0

⇒ xlog3 2 – (log223 + log2x) + 1 = 0

⇒ xlog3 2 – 3 – log2x + 1 = 0

⇒ xlog3 2 – 2 – log2x = 0

⇒2

2 xlog3

= (2 + log2x)2

⇒ 9 log2x = 4 + 4 log2x + log 22 x⇒ 9 log2x – 4 log2x = 4 + log2

2x⇒ 0 = log2

2x – 5 log2x + 4

Page 6: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution⇒ 0 = log2

2x – log2x – 4 log2x + 4⇒ 0 = log2x (log2x – 1) – 4(log2x – 1)∴ log2x = 4 = 16

log2x = 1 = 2

150. log2x – 3 log x = log (x2) = 4⇒ log2x – 3 log x – 2 log x + 4 = 0⇒ log10

2x – 5 log10x + 4 = 0⇒ log10

2x – log10x – 4 log10x + 4 = 0⇒ (log10x – 4) (log10x – 1) = 0⇒ log10x = 4 log10x = 1⇒ x = 104

⇒ x = 102

151. log1/3x – xlog3 3/1 + 2 = 0

⇒ – log3x – xlog3 3− + 2 = 0

⇒ (2 – log3x)2 = 2

3 xlog3

⇒ 4 + log32x – 4 log3x = – 9 log3x

⇒ log32 x + 5 log3x + 4 = 0

⇒ (log3x + 4) (log3x + 1) = 0⇒ log3x = – 4 log3x = – 1⇒ 3–4/3 = x 3–1 = x

⇒ x = 811

, 31

152. 2 2x )5(log – 3 logx 5 +1 = 0

⇒2

5 )x(log2 – xlog

3

5 + 1 = 0

⇒ xlog22

25

– xlog22

5 + 1 = 0

⇒ 0 = 2 log52x – 3 log5x + 1

⇒ 0 = 2 log5x(log5x – 1) – 1(log5x – 1)

⇒ log5x = 21

log5x = 1

⇒ x = 5 x = 5

153. xlog22 + 2 log2x – 2 = 0

⇒ xlog22 + log2x – 2 = 0

⇒ (log2x – 1) (log2x + 2) = 0⇒ log2x = 1 log2x = – 2⇒ x = 21 x = 2–2

Page 7: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

⇒ = 2 = 41

154. 2xlog )a( b – xlogbx5 + 6 = 0Let x logba = t

⇒ 2alog )x( b – alogbx5 + 6 = 0t2 – 5t + 6 = 0

⇒ t2 – 2t – 3t + 6 = 0⇒⇒ t(t – 2) – 3(t – 2) = 0⇒ (t – 3) (t – 2) = 0

⇒ t = 3 = alogbx

⇒ t = 2 =

alogbx

log 2 =

alogxlog

b

x = alogb2 x =

alogb3

155. log2 (100x) + log2 (10x) ± 14 + logx1

⇒ 2210 10log + xlog2

10 + 10log210 + xlog2

10 = 14 + log x1

⇒ 4 + log102 x + 1 + log102 x = 14 + log x1

156. log4 (x + 3) – log4 (x – 1)

⇒ log4 1x3x

−+

= 2 – 32 2log 2

⇒ log4 1x3x

−+

= 2 – 23

⇒ log4 1x3x

−+

= 21

⇒ 41/2 = 1x3x

−+

⇒ 2(x – 1) = (x + 3)⇒ 2x – 2 = x + 3⇒ x = 5

157. log4(x2 – 1) – log4(x – 1)2 = log4

2x4 −

⇒ log4 )1x)(1x()1x)(1x(

−−+−

= 21)x4(

42

log −

⇒ log4 1x1x

−+

= log4 (4 – x2)1/2

⇒2

1x1x

−+

= 2

2x4

Page 8: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

x21xx21x

2

2

−+

++ =

2x4−

158. 2 log3 7x3x

−−

+ 1 = log3 1x3x

−−

log3 2

7x3x

−−

+ 1 = log3

−−

1x3x

log3 2

7x3x

−−

– log3

−−

1x3x

= – 1

log3 2

2

)7x()3x(

− × 3x

1x−−

= – 1

2)7x()1x)(3x(

−−−

= 3–1

49x14x3x4x

2

2

+−

+− = 3

1

3x2 – 12x + 9 = x2 + 14x – 49⇒ 2x2 + 2x – 40 = 0⇒ x2 + x – 20 = 0⇒ x (x + 5) – 4(x + 5) = 0

(x – 4) (x + 5) = 0x = 4, – 5since 4 can’t satisfied the equation

∴ – 5 is a solution

159. 2 log4 (4 – x) = 4 – log2(–2 – x)

2 22log (4 – x) + log2(–2 – x) = 4

log2 (4 – x)(–2 – x) = 424 = x2 – 2x – 80 = x2 – 2x – 8 – 160 = (x – 6) (x + 4)∴ x = 6, –4

160. 3 + 2 logx + 1 3 = 2 log3(x + 1)Let log3(x + 1) = t

⇒ 3 + )1x(log2

3 + = 2 log3(x + 1)

⇒ = 3 + t2

= 2t

⇒ 3t + 2 = 2t2

⇒ 0 = 2t2 – 3t – 2⇒ 0 = 2t2 – 4t + t – 2⇒ 0 = 2t (t – 2) + l (t – 2)⇒ 0 = (2t + 1) (t – 2)

∴ log3(x + 1) = 21

− or log3x + 1 = 2

x + 1 = 3–1/2 x + 1 = 32

Page 9: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

x = 31

– 1 x = 8

2 –

+3

33

161. logx 9x2 . log32x = 4

(logx9 + logxx2) xlog2

3 = 4(logx3

2 + 2) log32x = 4

(2 logx 3 + 2) (log3

2x) = 4

xlog2

3 × log32x + 2 log3

2x – 4 = 0

2 log3x + 2 log32 x – 4 = 0

2 log32 x + 4 log3x – 2 log3x – 4 =0(2 log3x – 2) (log3x + 2) = 0log3x = –2 or log3x = 1

x = 91

x = 3

162. 22/1log (4x) + log2

8x2

= 8

– log22 4x + log2x2 – log28 = 8

– (log22 4 + log2

2x) + 2 log2x – 3 = 8– 2 log2

22 – log22x + 2 log2x – 11 = 0

163.10

x5.0log x2 – 14 log16xx3 + 40 log4x x = 0

164. 6 – )9.41( 3log24 3−+ . xlog7 = logx7

6 –

+ 9log4

4

399.41 log7x = xlog

1

7

6 –

+ 4

3 3log23

36.41 log7x = xlog1

7

6 –

+ 8336.41

165. log3 (4.3x – 1) = 2x + 132x + 1 = 4.3x – 132x . 31 = 4.3x – 1Let 3x = t

= 3t2 – 4t – 1 = 0 3x = 31

3t2 – 3t – t + 1 = 0(3t – 1) (t – 1) = 0 3x = 1

Page 10: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

t = 31

, 1 x = 0

166. log3 (3x – 6) = x – 1

3x – 1 = 3x – 6 3x = 9

33x

= 3x – 6 x = 22

3t

= t – 6

t = 3t – 18– 2t = – 18t = 9

167. log3 (4x – 3) + log3 (4

x – 1) = 1Let 4x = tlog3(4

x – 3) (4x – 1) = 1 Not possible31 = (t – 3) (t – 1) 4x = 43 = t2 – 4t + 3 x = 10 = t (t – 4)t = 0, 4

168. log3 (log1/22x – 3 log1/2x + 5) = 2

xlog22 1− – xlog3 12− + 5 = 32

⇒ – log22x + 3 log2x = 4

0 = log22x + 3 log2x + 4

0 = (log2x – 1) (log2

x + 4)∴ log2

x = 4 log2x = 4

x = 2 x = 16

169. log5 10x2 +

= 1x2

5log +

log5 10x2 +

– log5 1x2+

= 0

21x

1x2

5log+

×+ = 0

20x2xx2 2 +++

= 50

x2 + 3x + 2 = 20x2 + 3x – 18 = 0+ 6x – 3x(x + 6) (x – 3) = 0x = 3, – 6 Not satisfied3 Ans.

170. 1 + 2 logx + 25 = log5 x + 2Let log5x + 2 = t 5–1 = x + 2

x = 51

– 2 = – 59

Page 11: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

1 + 2xlog2

5 + = log5 x + 2 log5x + 2 = 2

1 + t2

= t 52 = x + 2

t + 2 = t2 x = 230 = t2 – t – 20 = (t – 2)–2 +1 (t + 1)t = –1, 2

171. log424x = 4log22

log424x = 424x = 44

44x = (22)4

24x = 84x = 8x = 2

172. log2

4x

= 18xlog

15

2 − Let log2x = t

⇒ log2x – log24 = 18logxlog15

22 −−

⇒ log2x – 2 = 13xlog15

2 −−

log2x – 2 = 4xlog15

2 −

(t – 2) = )4t(15−

(t – 2) (t – 4) = 15t2 – 2t – 4t + 8 = 15t2 – 6t + 8 – 15 = 0t2 – 6t – 7 = 0t2 – 7t + t – 7 = 0t (t – 7) + l(t – 7) = 0(t + 1) (t – 7) = 0t = –1, 7log2x = –1 or log2x = 7

x = 21

x = 27

173. 2

22

)x(log2xlog)x(log21

− =1 Let log x = t

xlog2xlog)xlog4(21

21010

210

− = 1 log10x = 0

1 – 8 log102x = log10x – 2 log10

2x x = 10°

0 = x2log106 + log10x – 1

Page 12: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution0 = (3 log10x – 1) (2 log10x + 1) = 0∴ 3 log10x = 1 or 2 log10x = – 1

log10x = 31

log10x = – 21

x = 3 10 x = 101

174. log2 (4.3x – 6) – log2 (9x – 6) = 1 Let 3x = t

log2(4.t – 6) – log2 (x2 – 6) = 1

log2 6t6t4

2 −

− = 1

4t – 6 = 2(t2 – 6)0 = 2t2 – 12 – 4t + 60 = 2t2 – 4t – 60 = t2 – 2t – 3t2 – 3t + t – 3 = 0∴ t (t – 3) + 1(t – 3) = 0

(t + 1) (t – 3) = 0∴ t = 3, – 1

3x = 3x = 13x = – 1 Rejected

∴ Answer = 1

175.21

log (5x – 4) + log 1x + = 2 + log 10018.0

21

log10 (5x – 4) + 21

log10x + 1 = 2 + log1018 – log10100

⇒21

log10 (5x – 4) (x – 1) = log1018

5x2 – 4x + 5x – 4 = 3245x2 + x – 328 = 0

D = b2 – 4ac = 1 + 6560 = 6561

D = 6561

x ≡ – 1082

, 8

∴∴∴∴∴ Final Ans. 8

176. log4 (2.4x – 2 – 1) + 4 = 2x

22log

−1

421.2 2

x

+ 4 = 2x

21

log2

−16

162.2 x2

+ 4 = 2x

log2

−+

16162 1x2

+ 4 = 4x

Page 13: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

16162 1x2 −+

177. logx 5 + logx 5x – 2.25 = (logx 5 )2

logx 5 + logx5 + logxx – 2.25 = 2

x 50log211

xlog21

5 + xlog

1

5 + 1 – 2.25 = xlog4

1

5Let log5x = t

t21

+ t1

+ 1 – 2.25 = t41

= t2)25.2(t2t221 −++

= t41

2 + 4 + 4t – 4t (2.25) = 16 + 4 (t – 2.25 t) = 1 – 6

–1 – 1.75 t = 45−

= 75

178. log (log x) + log (log x4 – 3) = 0log10 [log x + 4 logx – 3] = 05 log x – 3 = 15 log10x = 4

log10x = 54

Let logx = t

x = 104/5

⇒ 5 log2x – 3 log ? = 0log10t + 4 log2t – 4 log 3

179. log3x – 2 log1/3 x = 6log3x + 2log3x = 6log3x + log3x

2 = 6x3 = 36

x3 = (32)3

x = 9

180. )4x5log(xlog2− = 1

⇒ 2 log x = log (5x – 4)log x2 = log (5x – 4)x2 – 5x + 4 = 0x2 – 4x – x + 4 = 0x(x – 4) –1(x – 4) = 0(x – 1) (x – 2) = 0x = 1, 4

Answer = 4

181. 2 log82x + log8x2 + 1 – 2x =

log8(2x)2 + log8 x2 – 2x + 1 = 3

4

Page 14: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

log84x2 (x2 – 2x + 1) = 34

4x2(x2 – 2x + 1) = (23)4x2(x2 – 2x + 1) = 24

4x2 (x2 – 2x + 1) = 16x2 [x2 + 2x + 1] = 4x4 + 2x3 + x2 – 4 = 0

2 log82x + log8(x2 + 1 – 2x) = 3

4

2[log82 + log8x] + log8 x2 – 2x + 1 = 3

4

(x + 1) (x3 – 3x2 + 4x – 4)(x – 2) (x2 – x + 2)Thus (x – 1) (x – 2) (x2 – x + 2) are its factorsince 1 can’t satisfies, 2 satisfies the answer∴ Answer = 2

182. 61

log2(x – 2) – 31

= log1/8 5x3 −

61

log2 (x – 2) = 31

= 32log − 5x3 −

61

log2 (x – 2) – 31

= – 31

log2(3x – 5)1/2

61

log2 (x – 2)1/6 . (3x – 5)1/6 = 31

= log2 (x – 2)1/6 . (3x – 5)1/6 = 31

((x – 2)(3x – 5))1/6 = (2)1/3 × 6

if 9 multiplied power by, get3x2 – 6x –5x + 10 = 43x2 – 11x + 6 = 03x (x – 3) – 2(x – 3) = 0

x = 32

, 3

Since 32

doesn’t satisfic equation

∴ final answer = 3

183. 2 log3(x – 2) + log3 (x – 4)2 = 0log3(x – 2)2 + log3 (x – 4)2 = 02 log3(x – 2) + 2 log3 (x – 4) = 02[log3(x – 2) (x – 4)] = 0x2 – 2x – 4x + 8 = 3° D = b2 – 4axx2 – 6x + 8 – 1 = 0 36 – 28 = 8

x2 – 6x + 7 = 0 D = 2 2

Page 15: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

∴ x = 2

226 ±

= 2 2

)23( ±

= 3 + 2 , 3 – 2

184. x16log).x2(log 42

2 = log4x3 let log2x = t

)xlog16)(logxlog2(log 442

22 ++ = 3 log4x

++ xlog212)xlog21( 22 =

23

log2x

=

2

2t2)t21(

++ = 2

t23

(1 + 2t)

+2

t4 =

4t9 2

2 (4 + 8t + t + 2t2) = 9t20 = 9t2 – 4t2 – 18t – 80 = 5t2 – 18t – 80 = 5t2 – 20t + 2t – 80 = 5t (t – 4) + 2(t – 4)0 = (5t + 2) (t – 4)

∴ t = 4, – 52

∴ log2x = 4 log2x = – 52

x = 24 x = (2)–2/5 can’t satisfied mex = 16

185. 1xlog319xlog3−+

= 2 logx + 1 Let log10x = t

1t319t3−+

= 2t + 1 log10x = 2

3t + 19 = (2t + 1)(3t – 1) log10x = – 35

3t + 19 = 6t2 + 3t – 2t –1 x = 10– 5/3

0 = 6t2 – 2t – 200 = 3t2 – t – 100 = 3t2 – 6t + 5t – 100 = (3t + 5) (9t – 2)

186. 40xlog)11xlog(

3 −

++ = 3

Page 16: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

log )11x( ++ = 3

− 40xlog3

log 1x + + 1 = logx – 120

120 = logx – log 1x + – 1

187. log326 – log3

2 2 = (log2x – 2) log3121log3

2 2 + log323 – log3

22 = (log2x – 2) log312(log3x)2 = (log10

2x – 2) (log33 + log34)

188. 1 – 21

log (2x – 1) = 21

log (x – 9)

1 = 21

log (x – 9) + 21

log (2x – 1)

1 = 21

[log (x – 9) (2x – 1)]

2 = log10 2x2 – 19x + 9102 = 2x2 – 19x + 90 = 2x2 – 19x – 910 = 2x2 – 26x + 7x – 910 = 2x (x – 13) + 7(x – 13)0 = (2x + 7) (x – 13)

x = – 27

, 13

x = 13 Ans.

189. )5xlog(8log2log7xlog

−−−+

= – 1

5x8

10

27x

10

log

log

+

= – 1

log10 27x +

= ? – log10 5x8−

190. log(3x2 + 7) = log (3x – 2) = 1log (3x2 + 7) = 1 log (3x – 2) = 1101 = 3x2 + 7 101 = 3x – 23 = 3x2 12 = 3xx = 1 x = 4log (3x2 + 7) – log (3x – 2) = 0

log 2x37x3 2

−+

= 0

3x2 + 7 = 3x – 23x2 – 3x + 9 = 0x2 – x + 3 = 0

191. 31

log(x2 – 16x + 20) – log 3 7 = 31

log (8 – x)

Page 17: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

31

[log10x2 – 16x + 20 – log107] = 3

1 log108 – x

log10 720x16x2 +−

= log108 – x

x2 – 16x + 20 = 56 – 7xx2 – 16x + 20 – 56 + 7x = 0x2 – 9x – 36 = 0– 12 + 3x(x – 12) + 3(x – 12) = 0(x + 3) (x – 12) = 0∴ x = –3, 12makes the x < 1 rejectedx = –3 Ans.

192. log63x + 3 – log6 3

x – 2 = x

log6 2x

3x

32

+

= x

2x

3x

32

+

= 6x

2x

3x

32

+

= 2x . 3x

2

x

x

x

3332

= 82x

× x39

= 2x .3x

193.

+x2

11 log 3 + log 2 = log (27 – 2 3 )

194. log(5x – 2 + 1) = x + log 13 – 2 log 5 + (1 – x) log 2

196. log2(4x + 1) = x log2(2

x + 3 – 6)

log2 (22x + 1) = x + log2

− 6

82x

198. log3 (9x + 9) = x + log3 (28 – 2 . 3x)

199. log (log x) + log (log x3 – 2) = 0 Pattern may be same of Q.No. 178log10 [log x + logx3 – 2] = 04 log x – 2 = 1

log x = 43

x = 103/4

Answer has been given is 110.

Page 18: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

200. 5log 4x – 6 = – 5log 2x– 1 = 2

5log2264

x

x

− = 2 Let 2x = t

2 log5 2t6t2

−−

= 2 2x = 4

log5 2t6t2

−−

= 1 2x = 4 x = 6

2t6t2

−−

= 5a1 ∴ x = 2

t2 – 6 = 5t – 10t2 – 5t + 4 = 0t(t – 1) – 4(t – 1) = 0t = 4, 1

203. log

− x

4x

223 = 2 +

41

log 16 – 24log x

204.xlogxlog2

51 −

= 125

1. 5logx – 1

205.xlog

32xlog3 2

x−

= 3 10100

206. log2 (25x + 3 – 1) = 2 + log2 (5 x + 3 + 1)

210. log2(2x2) . log2 (16x) = 29

log22x

(log22 + log2x

2) (4 + log24) = 29

log22 x

(1 + 2 log2x) (4 + log2x) = 29

log22x

⇒ 2 [4 + 8 log2x + log2x + 2log22x] = 9 log2

2x⇒ 8 + 16 log2 x + 2 log2x + 4 log2

2x = 9 log22 x

0 = 5 log22x – 18 log2x – 8 Let log2x = t

0 = 5t2 – 18t – 8tD = b2 – 4ac= 324 + 160= 484

D = 22

t = 102218 ±

= 1040

= 4

∴ log2x = 4 log2x = – 52

x = 24 x = (2)– 2/5

Page 19: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution16 = 2 – 2/5

211. log44 +

+x2

11 log3 = log

+ 2733

since radical should be defined

log44 + 2411

3log+

= log

+ 273x

log (4.3 . 31/2x) = log (31/x + 27) Let 31/x = t12.t = t2 + 27 31/x = t0 = t2 – 12t + 270 = t2 – 9t – 3t + 270 = (t – 9) (t – 3)∴ t = 9, 3

x = 1, 21

212. log(x3 + 27) – log (x2 + 6x + 9) = 7log 3

log (x + 3) (x2 – 3x + 9) – log 2

2/12

)3x()9x6x(

+

++ = log 7

log10

++−+

)3x()9x3x)(3x( 2

= log107

x2 – 3x + 9 – 7 = 0x2 – 3x + 2 = 0x2 – 2x – x + 2 = 0x(x – 2) – 1(x – 2) = 0(x – 1) (x – 2) = 0x = 1, 2

213. 5log x = 50 – xlog 5

5log x + xlog 5 = 50 Let log10x = t5log x + 5log x = 50 x = 102 = 1005t + 5t = 502.5t = 505t = 25t = 2

215. xlog1xlog3

x−

= 3 10

xlog1xlog3 2

x−

= 3 10

log xlog1xlog3 2

x−

= log 3 10

log x . xlog1xlog3 2 −

= 31

log 10

3 log2x – 1 = 31

Page 20: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

3 log2x = 34

log102x = 9

4

log10x = ± 3

2

x = (10)2/3 , (10–2/3

216. |x – 10| log2 (x – 3) = 2(x – 10)log2 (x – 3)|x – 10| = 2(x – 10)(x + 10) log2(x – 3) = 2(x – 10)1 – (x + 10) log2(x – 3) = 2(x – 10)log2(x – 3) = 2 log2(x – 3) = – 2x – 3 = 29 (x – 3) = 2–2

x = 7 (x – 3) = 41 ⇒ x =

41

+ 3 = 4

13

217. log4 log2x + log2 log4x = 2 log2x = t

21

log2 log2x + log2 21

log2x = 2 2x = 4

21

log2t + log 2 2t

= 2 x = 24

log2 t1/2 + log2t – log22 = 2log2t

1/2 + log2t – log22 = 2log2t

1/2 + log2t = 3

21

log2t + log2t = 3

log2t + 2 log2t = 63 log2 t = 6t = 22 = 4

218. (6x – 5) |ln (2x + 2.3)| = 8 ln (2x + 2.3)

(6x – 5) = |)3.2x2(n|)3.2x2(n8

++

l

l

Ist 6x – 5 = – 86x = 13

x = 613

= – 21

It is not in answer

IInd 6x – 5 = 8 The right answer

6x = 13 = 613

, 2013−

x = 613

219. x3logx9log 38

9 − = log3x3

= )xlog3.(logxlog9log 338

99 ++ = 3 log3x

Page 21: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

=

2

33 )xlog1.(xlog281

++ = (3 log3x)2

= (1 + 4 log3x) (1 + log3x) = 9 log32 x

(1 + 4t) (1 + t) = 9t20 = 9t2 – 4t2 – 5t – 10 = 5t2 – 5t – 1D = b2 – 4ac = 25 + 20 = 45

t = 10

535 ±

log3x = t

log3 x = 10

535 ±

x = 3 10

535 ±

220. log2 (100x) – log2 10x + log2x = 6

log102 102 + log10

2x – ]xlog10[log 210

210 + + log10

2x = 6

4 + xlog1log 210

210 −− = 6

210log x = 3

log10x = 3

221. )1x(log 3/19 + = )1x2(log 25/15 +

)1x(log2 133+− =

)1x2(log 2155

+−

23 )1x(log23

−+− = (2x2 + 1)–1

(x + 1)–2 = (2x2 + 1)–1

2)1x(1+ =

1x212 +

2x2 + 1 = x2 + 1 + 2x2x2 – x2 – 2x = 0x2 – 2x = 0x(x – 2) = 0x = 0, 2

223. 2 log2

−−

1x7x

+ log2

+−

1x1x

= 1

log2 )1x)(1x(x1449x2

−−−+

× 1x1x

+−

= 1

⇒1x

x14490x2

2

−+ = 2

⇒ x2 + 49 – 14x = 2x2 – 2⇒ x2 + 14x – 2 – 49 = 0

x2 + 14x – 51 = 0x2 + 17x – 3x – 51 = 0

Page 22: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solutionx(x + 17) – 3(x + 17) = 0

∴ x = – 17, 3→can’t satisfy the equation.∴ x = – 17 only

225.21

log (1 + x) +23

log (1 – x) = 21

log (1 – x2)

= 21

[log (1 + x) + log (1 – x)3] = 21

log (1 – x2)

log )x1()x1(x1

2 −−+

= log (1 – x2)

⇒ log )x1()x1(x1

2 −−+

– log(1 – x2) = 0

2

3

10 x1)x1(log

x1

−−

+ = 0

3)x1()x1(

−+

× )x1)(x1(1

−+ = 1

log10 4)x1(1− = 0

log10 (1 – x)–4 = 0– 4 log10 (1 – x) = 0100 = 1 – xx = 0

226. )x5log()x35log( 3

−−

= 3

log (35 – x3) = 3 log (5 – x)log (35 – x3) = log(5 – x)3

35 – x3 = 125 – x3 – 15x (5 – x)35 – 125 = – 75x + 15x2

0 = 15x2 – 75 x + 900 = x2 – 5x + 60 = x2 – 2x – 3x + 6∴ x = 2, 3

227. logx2 – log4x + 67

= 0

xlog1

2 –

21

log2x + 67

= 0

6 – 3 log22 x ± 7 log2x = 0

3 log22 x – 9 log2x + 2 log2x – 6 = 0

(3 log2x + 2) (log2x – 3) = 03 log2x = – 2 log2x = 3

log2x = – 32

x = 23

x = 2 – 2/3 x = 8

Page 23: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

229. xlog2 – 0.5 = log2 x

xlog2 = 21

log2x + 21

xlog2 = 21

[log2x + 1]

4 log2x = log22x + 1 + 2 log2x

0 = log22 x – 2 log2x + 1

0 = log2x (log2x – 1) – 1(log2x – 1)0 = log2x – 11 = log2xx = 2

230. log1/3

× 1212

x

= log1/3

421 x2

2 x

21

– 1 =

x2

21

– 4 Let

x

21

= t

2 x

21

21

. x

21

= – 3

⇒ 2t – t2 = 3⇒ 0 = t2 – 2t – 3⇒ 0 = t(t + 1) – 3(t + 1)⇒ (t – 3) (t + 1) = 0⇒ t = 3, –1

Q

x

21

= t

x

21

= – 1, 3 –1 not defined

log x

21

= log 3

log 2–x = log 3– x log 2 = log 3

– x = 2log3log

x = – log23 Ans.

231.21

log6(x – 2) + 21

log6(x – 11) = 1

21

log6 (x – 2) (x – 11) = 1

log6 x2 – 13x + 22 = 2

x2 – 13x + 22 = 62

x2 – 13x + 22 – 36 = 0x2 – 13x – 14 = 0x2 – 14x + x – 14 = 0x(x – 14) + 1(x – 14) = 0

Page 24: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution(x + 1)(x – 14) = 0x = 14 Ans.–1 is not defined

233. log7 2 + 27log x = 2/17 )3(log 1−

log72 + 21

log7x = – 21

log73

log72 + 21

log73 + 21

log7x = 0

2 log72 + log7 3 + log7x = 0log7 12 = – log7xx = – 12

237. log3x x3

+ log32x = 1 Let log3x = t

x3logx3log

3

3 + log3

2 x = 1 log3x = 0

xlog3logxlog3log

33

33

+−

+ log 32x = 1 log3x = 1

t1t1

+−

+ t2 = 1 x = 1

1 – t + t2 (1 + t) = 1 + t x = 3–2

t3 + t2 – 2t = 0t(t2 + t – 2) = 0t2 + t – 2 = 0t(t + 2) – 1(t + 2) = 0t = 1, 2Ans. 1, 3, 3–2

246. log4(x + 12) logx2 = 1

21

log xlog12x

2

+ = 1

log2 (x + 12) = 2 log2xlog2 x + 12 = log2x

2

x + 12 = x2

0 = x2 – x – 120 = x2 – 4x + 3x – 120 = x(x – 4) + 3(x – 4)x = 4, 3 not satisfied∴ 24 Ans.

247. log16x + log4x + log2x = 7

41

log2x + 21

log2x + log2x = 7

4xlog4xlog2xlog 22 ++

= 7

7 log2x = 28log2x = 4

Page 25: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solutionx = 16

248. log 10 + 31

log )2713( x2 + = 2

31

log )2713( x2 + = 1

log10 )2713( x2 + = 3

x23 + 271 = 103

x23 = 1000 – 271

x23 = 279

x23 = 36 ⇒ x2 = 6

x = 3 ⇒ 2)x( = 9 ⇒ x = 9

250. log5 (3x + 10) + 7.10 = log5 (9

x + 156)Detective Question for me and how can I solve

251. (log4x – 2) log4x = 23

(log4x – 1)

2(log4x – 2) log4x = 3 log4x – 3 Let log4x = t(2t – 4) t = 3t – 32t2 – 4t – 3t + 3 = 02t2 – 7t + 3 = 02t2 – 6t + t + 3 = 02t (t – 3) – 1(t – 3) = 0

t = 3, 21

∴ log4x = 3, 21

x = 43 , (4)1/2

x = 64, 2

252. log5x + log25x = log1/5 3

log5x + 21

log5x = – 21

log5 3

2 log5x + log5x = – log 53

– 3 log5x = log53log5x

–3 = log5 3(x–3)1/3 = (3)– 1/3

x = 3 31

254. log2 (9 – 2x) = 3 – x23 – x = 9 – 2x Let 2x = t

Page 26: Log Equations

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

Mathematics Online: It’s all about Mathematics A. I. Prilepko Solution

x

3

22

= 9 – 2x 2x = 1

t8

= 9 – t 2x = 8

8 = 9t – t2 x = 3t2 – 9t + 8 = 0t2 – 8t – t + 8 = 0t (t – 8) – 1 (t – 8) = 0(t – 1) (t – 8) = 0t = 1, 8x = 0, x = 3 Ans.

255. 2(log 2 – 1) + log )15( x + + log )55( x1 +−

2(log 2 – 1) + log )15( x + + log

+ 5

5

5x

256. logx3 + log3x = 3log x + xlog3 + 21

xlog1

3 + log3x = xlog

2

3 +

21

log3x + 21

t1

+ t = t2

+ 2t

+ 21

tt1 2+

= t2

tt2 2 ++

2 + 2t2 = 2 + t2 + tt2 – t = 0 Q log3x = 0t = 0, 1 x = 30 = 1

log3x = 1 ⇒ x = 3

259. logx 125x . xlog225 = 1 2

5 )x(log 2

xlogx125log

5

5 . log5

22x = 12

5 xlog21

xlogxlog5log

5

53

5 + .

21

log25x = 1

tt3 +

. 4t2

= 1

t4tt3 32 +

= 1

t(3t + t2) = 4tt2 + 3t – 4 = 0t2 + 4t – t – 4 = 0t(t + 4) – 1(t + 4) = 0(t – 1) (t + 4) = 0t = 1, – 4∴ log5x = 1, –4

x = 5, 5–4