log equations
TRANSCRIPT
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
}
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
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±
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
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
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
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
−
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
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
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
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
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
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
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
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
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)
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.
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
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
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
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
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
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
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
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
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