Log Equations

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}

⇒

x2 _{+ 1 – 2x = 3}

⇒

x2 _{– 2x – 2 = 0} D = b2 _{– 4ac = 12} Roots = −b₂+_a D = 2 3 2 2± ₌ 2 ) 3 1 ( 2 ± = 1 ± 3

Since 1+ 3satisfies the equation ∴∴∴∴∴ Final solution x ∈{

3 1+

}

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

2

1

_⇒

2 1

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

2 1

⇒

2 log₃ (1 + log₂(1 + 3 log₃x))) = 21 _{= 2}

⇒

_{(1 + log}

2 (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 + log}

3 (2x – 7) = 3

⇒

(2x – 7) = 32

⇒

2x _{= 9 + 7 = 16}

⇒

_{x = 4 ∴∴∴∴∴ Final solution x} ∈ {4} 128. log₃ (3x _{– 8) = 2 – x}

⇒

₃2 – x _{= 3}x _{– 8}

⇒

x 2 3 3 = 3x _{– 8}

⇒

x 3 9 = 3x _{– 8} Let 3x _{= t} t 9 = t – 8

⇒

9 = t2 _{– 8t}

⇒

_t2 _{– 8t – 9 = 0} D = b2 _{– 4ac = 64 + 36 = 100} t = 3x ₌ 2 10 8± = 2 18 , – 2 2

= 9 ,–1 (negative value is not possible as ax _{>0 )} 3x ₌₉

⇒

₃x _{= 3}2

⇒

_{x = 2 ∴∴∴∴∴ Final solution x ∈{2}} 129. x 3 ) 2 9 ( log₂ x − − = 1 x ≠ 3

⇒

log₂ (9 – 2x_{) = 3 – x}

⇒

_{9 – 2}x _{= (2)}3 – x

⇒

_{9 – 2}x ₌ x 3 2 2

⇒

8_t = 9 – t Now, 2x _{= 1, 8}

⇒

_{8 = 9t – t}2

⇒

_t2 _{– 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} (x2 _{– 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. log₃ (log₉x + 2 1 + 9x_{) = 2x}

⇒

_log 3 (log9x + ₂ 1 + 32x_{) = 2x}

⇒

_– 2 1 = 2 1 log₃x –1 = log₃x

⇒

x = 3 1 ∴ ∴∴ ∴∴ Final solution x ∈{ 3 1 }

132. log₃(x + 1) (x + 3) = 1

⇒

31 _{= (x + 1) (x + 3)}

⇒

_{3 = x}2 _{+ x + 3x + 3} 3 = x2 _{+ 4x + 3}

⇒

_{0 = x}2 _{+ 4x}

⇒

_{0 = x(x + 4)}

∴ x = –4, 0 since – 4 doesn’t satisfy equation ∴∴∴∴∴ Final solution x ∈{0} 133. log₇ (2x _{– 1) + log}

7 (2x – 7) = 1 Let 2x _{= t}

log₇(t – 1) (t – 7) = 1

⇒

71 _{= t}2 _{– 7t – t + 7}

⇒

_{0 = t}2 _{– 8t}

⇒

_{t = 0, 8} 2x _{= 8}

⇒

_{x = 3}

2x _{= 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

1 x 2 20 x 21 − − ⇒ log 5(x₁₀+10) = log 1 x 2 20 x 21 − − ⇒x+₂10 = 1 x 2 20 x 21 − − ⇒ (2x – 1) (x + 10) = 42x – 40 ⇒ 2x2 _{– x + 20x – 10 = 42x – 40} ⇒ 2x2 _{– 19x – 42x – 10 + 40 = 0 ⇒ 2x}2 _{– 20x – 3x + 30 = 0} ⇒ (2x – 3) (x – 10) = 0 ⇒ x = 2 3 , 10 ∴ ∴∴ ∴∴ Final solution x ∈{ 2 3 , 10 } 135. 1 – log5 = ₃1       _{+ + log5} 3 1 x log 2 1 log _{⇒ log} 10 10 – log105 = ₃ 1 [log₁₀2–1 _{+ log} 10x + log1051/3] ⇒ log₁₀ 5 10 = 3 1 [log₁₀2–1 _{+ log} 10x + log1051/3] ⇒ log1023 = ₃ 1 log₁₀2–1 _{+ log} 10x + log1051/3 ⇒ log₁₀23 _{= log} 10     _x 3₅ 2 1 ⇒ 8 = 2 5 x3 ⇒ x = 3₅ 16 ∴ ∴∴ ∴∴ Final solution x ∈{ 3₅ 16 } 136. log x – 2 1 log       − 2 1 x _{= log}       + 2 1 x _– 2 1 log       + 8 1 x _{⇒log x – log} 2 1 x− = log x + 2 1 – log x+₈1 ⇒ log 2 1 x x − = log 8 1 x 2 1 x + + ⇒ 2 2 1 x x             − = 2 8 1 x 2 1 x             + + ⇒ 2 1 x x2 − = 8 1 x x 4 1 x2 + + + ⇒ x3 ₊ 8 x2 = x3 ₊ 4 x + x2 _– 2 x2 – ₈1 – 2 x ⇒ 8(8 x3 _{+ x}2_{) = 8(8x}3 _{+ 2x + 8x}2 _{– 4x}2 _{– 1 – 4x)} ⇒ 8x3 _{+ x}2 _{= 8x}3 _{– 2x + 4x}2 _{– 1 ⇒ 0 = 3x}2 _{– 2x – 1 ⇒ 0 = 3x}2 _{– 3x + x – 1} ⇒ 0 = 3x (x –1) +1(x – 1) ⇒ x = – 3 1 , 1 ∴∴∴∴∴ Final solution x ∈{ – 3 1 } 137. ₃log3log x _{– logx + log}2_{x – 3 = 0} ⇒ _{log x – log x + log}2_{x – 3 = 0}

⇒ log₁₀ x – log₁₀x + log₁₀2x – 3 = 0 ⇒ ₂1 log₁₀x – log₁₀x + (log₁₀x)2 _{– 3 = 0} ⇒ log₁₀x – 2 log₁₀x + 2(log₁₀x)2 _{– 62 = 0 ⇒ 2(log}

10x)2 – log10x – 6 = 0 ⇒ 2(log₁₀x)2 _{– 4 log}

10x + 3 log10x – 6 = 0 ⇒ (2 log10x + 3) (log10x – 2) = 0 ⇒ log₁₀x = 2 or log₁₀x = –

2 3

138. (x−2)log2(x−2)+log(x−2)5−12= 102 log (x – 2) _{(Let log (x-2) = t )} ⇒ log (x−2)log2(x−2)+log(x−2)(5−12) = log 102log (x – 2)

⇒ log (x – 2) (log2_{(x – 2)+ log (x – 2)}5 _{– 12)) = log}

10102log(x – 2) ⇒ t [t2 _{+ 5t – 12)] = 2t ⇒ t}3 _{+ 5t}2 _{– 14t = 0 ⇒ t(t}2 _{+ 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. 9log3(1−2x) _{= 5x}2 _{– 5}

⇒

32log3(1−2x)_{= 5x}2 _{– 5}

⇒

2 3(12x) log 3 − = 5x2 _{– 5}

⇒

(1 – 2x)2 _{= 5x}2 _{– 5}

⇒

1 + 4x2 _{– 4x = 5x}2 _{– 5}

⇒

0 = 5x2 _{– 4x}2 _{– 4x – 5 – 1}

⇒

0 = x2 _{– 4x – 6} D = b2 _{– 4ac} = 16 + 24 = 40 D= 2 10 α = a 2 D b+ − = 2 + 10 , 2 – 10 140. x1 + log x _{= 10x} log_xx1 + logx_{= log 10x}

⇒

(log₁₀x) (1 + log₁₀x) = log₁₀10 + log₁₀x ⇒ log₁₀x + log₁₀2_{x = 1 + log}

10x

⇒

log₁₀2_{x = 1}

⇒

log₁₀ x = ±1

∴ x = 10–1_{, 10}1

141. x2 logx _{= 10x}2

⇒

log₁₀x2_{.log x = log}

1010 + log10x2

⇒

2 log₁₀x (log₁₀x) = 1 + 2 log₁₀x

⇒

2 log₁₀2_{x – 2 log} 10x – 1 = 0 Let log10x = t

⇒

2t2 _{– 2t – 1 = 0} 2t2 D = b2 _{– 4ac} = 4 + 8 = 12 D = 2 3 t = 9 3 2 2± = 2(1±₄ 3) = 2 3 1± Now, log₁₀x = t = 2 3 1+ , 2 3 1− ∴ x = 2 3 1±

⇒

log₁₀x = 2 3 1±

⇒

log₁₀x2 _{= 1± 3}

⇒

x2 _{= 10}1± 3 x = ₁₀1+ 3 _{, 10}1− 3 142. _xlog₃x+5 = 105 + log x log x . 3 5 x log + = log₁₀105 _{+ log} 10x ⇒ logx. 3 5 x log + _{= 5 + log} 10x ⇒ log2_{x + 5 log} 10x = 15 + 3 log10x ⇒ log₁₀2_{x + 2 log} 10x – 15 = 0

⇒ log₁₀x (log₁₀x + 5) – 3(log₁₀x + 5) = 0

⇒

(log₁₀x – 3) (log₁₀x + 5) = 0

⇒

log₁₀x = 3 log₁₀x = – 5

⇒

x = 103 _{x = 10}–5 143. xlog3.x _{= 9}

⇒

log3.x 3x log = log₃9

⇒

log₃x (log₃x) = 21

⇒

log₃2_{x = 21}

⇒

log₃x = ± ₂

⇒

x = ₃ 2_{, 3}− 2 144. _{( x)}log5x−1 _{= 5}

⇒

log₅ x. log5 (x – 1) = log55

⇒

₂1log₅x (log₅x – 1) = 1

⇒

₂1log₅2_{x –} 2 1 log₅x = 1

⇒

= log₅2 _{x – log} 5x = 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 =}

5 1

⇒

x = 5 and x = ₅1

145. xlogx +1 _{= 10}6

⇒

log x (log x + 1) = log₁₀106

⇒

log₁₀2_{x + log}

10x – 6 = 0

⇒

(log₁₀x – 2) (log₁₀x + 3) = 0

⇒

log₁₀x = 2 log₁₀x = – 3

⇒

x = 10 x = 10 146. 4 7 x log x + = 10log x + 1

⇒

log x. log 4 7 x+ = 4 log₁₀ x + 4

⇒

log2 x 10 + 3 log10x – 4 = 0

⇒

log2 x 10 + 4 log10x – log10x – 4 = 0

⇒

(log₁₀x + 4) (log₁₀x – 1) = 0

⇒

x = 104 _{and x = 10}1 147. _xlog x(x−2) _{= 9}

⇒

x1/2logx(x−2) = 9

⇒

2 x(x 2) log x − = 9

⇒

(x−2)2logxxx_{= 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. 2 x log x log2 2 2 x log + −       = log x

⇒

₂1 log2x logx2 2 x log + − = 2 1 2 log x

⇒

log x logx 2 2 2 2 x log + −       = log x

⇒

log x logx 2 2 2 2 x log + −       = 2 log x 149. 3 log2x – log28x + 1 = 0

⇒

3 log₂x – (log₂23 _{+ log}

2x) + 1 = 0

⇒

3 log₂x – 3 – log₂x + 1 = 0

⇒

3 log₂x – 2 – log₂x = 0

⇒

_3 log2x _2 = (2 + log₂x)2

⇒

9 log₂x = 4 + 4 log₂x + log 2₂ x ⇒ 9 log₂x – 4 log₂x = 4 + log₂2_x

⇒

0 = log₂2_{x – 5 log}

⇒

0 = log₂2_{x – log}

2x – 4 log2x + 4

⇒

0 = log₂x (log₂x – 1) – 4(log₂x – 1) ∴ log₂x = 4 = 16

log₂x = 1 = 2

150. log2_{x – 3 log x = log (x}2_{) = 4}

⇒

log2_{x – 3 log x – 2 log x + 4 = 0}

⇒

log₁₀2_{x – 5 log} 10x + 4 = 0

⇒

log₁₀2_{x – log} 10x – 4 log10x + 4 = 0

⇒

(log₁₀x – 4) (log₁₀x – 1) = 0

⇒

log₁₀x = 4 log₁₀x = 1

⇒

x = 104

⇒

x = 102 151. log_1/3x – 3 log_1/3x + 2 = 0

⇒

– log₃x – 3 −log₃x + 2 = 0

⇒

(2 – log₃x)2 ₌ 2 3x log 3       ₋

⇒

4 + log₃2_{x – 4 log} 3x = – 9 log3x

⇒

log₃2 _{x + 5 log} 3x + 4 = 0

⇒

(log₃x + 4) (log₃x + 1) = 0

⇒

log₃x = – 4 log₃x = – 1

⇒

3–4/3 _{= x 3}–1 _{= x}

⇒

x = 81 1 , 3 1 152. 2 2 x 5) (log – 3 log_x 5 +1 = 0

⇒

₂ 5x) (log 2 _– x log 3 5 + 1 = 0

⇒

_2log22_x 5 – 2log x 2 5 + 1 = 0

⇒

0 = 2 log₅2_{x – 3 log} 5x + 1

⇒

0 = 2 log₅x(log₅x – 1) – 1(log₅x – 1)

⇒

log₅x = 2 1 log₅x = 1

⇒

x = 5 x = 5 153. log2x 2 + 2 log2x – 2 = 0

⇒

log2₂x+ log₂x – 2 = 0

⇒

(log₂x – 1) (log₂x + 2) = 0

⇒

log₂x = 1 log₂x = – 2

⇒

x = 21 _{x = 2}–2

⇒

= 2 = 4 1 154. (alogbx)2 _{– 5x}logbx + 6 = 0 Let x log_ba = t

⇒

_(xlogba₎2 _{– 5x}logba + 6 = 0 t2 _{– 5t + 6 = 0}

⇒

t2 _{– 2t – 3t + 6 = 0}

⇒⇒

t(t – 2) – 3(t – 2) = 0

⇒

(t – 3) (t – 2) = 0

⇒

t = 3 = _xlogba

⇒

t = 2 = a logb x log 2 = a log x log b x = ₂logba _{x =} a logb 3

155. log2 _{(100x) + log}2 _{(10x) ± 14 + log}

x 1

⇒

2 2

1010

log + log₁₀2 x + log₁₀2 10 + log₁₀2 x = 14 + log

x 1

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

x 1 156. log₄ (x + 3) – log₄ (x – 1)

⇒

log₄ 1 x 3 x − + = 2 – 3 2 2 log 2

⇒

log₄ 1 x 3 x − + = 2 – 2 3

⇒

log₄ 1 x 3 x − + = 2 1

⇒

41/2 ₌ 1 x 3 x − +

⇒

2(x – 1) = (x + 3)

⇒

2x – 2 = x + 3

⇒

x = 5 157. log₄(x2 _{– 1) – log} 4(x – 1)2 = log4 4−x2 ⇒ log_{4 (}(x_x−₋1₁₎₍)(_xx₋+1₁₎) = 2 1 ) x 4 ( 4 2 log −

⇒

log₄ 1 x 1 x − + = log₄ (4 – x2₎1/2

⇒

2 1 x 1 x       − + = 2 2 x 4 _    ₋

x 2 1 x x 2 1 x 2 2 − + + + = 4−x2 158. 2 log₃ 7 x 3 x − − + 1 = log₃ 1 x 3 x − − log₃ 2 7 x 3 x       − − + 1 = log₃ _x_x−₋3₁_ log₃ 2 7 x 3 x       − − – log₃ _x_x−₋3₁_ = – 1 log_{3 2} 2 ) 7 x ( ) 3 x ( − − × 3 x 1 x − − = – 1 2 ) 7 x ( ) 1 x )( 3 x ( − − − = 3–1 49 x 14 x 3 x 4 x 2 2 + − + − = ₃1 3x2 _{– 12x + 9 = x}2 _{+ 14x – 49} ⇒ 2x2 _{+ 2x – 40 = 0} ⇒ x2 _{+ x – 20 = 0} ⇒ x (x + 5) – 4(x + 5) = 0 (x – 4) (x + 5) = 0 x = 4, – 5

since 4 can’t satisfied the equation ∴ – 5 is a solution 159. 2 log₄ (4 – x) = 4 – log₂(–2 – x) 2 log (4 – x) + log₂2 ₂(–2 – x) = 4 log₂ (4 – x)(–2 – x) = 4 24 _{= x}2 _{– 2x – 8} 0 = x2 _{– 2x – 8 – 16} 0 = (x – 6) (x + 4) ∴ x = 6, –4 160. 3 + 2 log_{x + 1} 3 = 2 log₃(x + 1) Let log₃(x + 1) = t

⇒

3 + _{log (}2_{x 1)} 3 + = 2 log3(x + 1)

⇒

= 3 + 2_t = 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) = 2 1 − or log₃x + 1 = 2 x + 1 = 3–1/2 _{x + 1 = 3}2

x = 3 1 – 1 x = 8 2– _       ₊ 3 3 3 161. log_x 9x2 _{. log} 32x = 4 (log_x9 + log_xx2_{) log}2_x

3 = 4 (log_x32 _{+ 2) log} 32x = 4 (2 log_x3 + 2) (log₃2_{x) = 4} x log 2 3 × log32x + 2 log3 2_{x – 4 = 0} 2 log₃x + 2 log₃2 _{x – 4 = 0}

2 log₃2 x + 4 log₃x – 2 log₃x – 4 =0 (2 log₃x – 2) (log₃x + 2) = 0 log₃x = –2 or log₃x = 1 x = ₉1 x = 3 162. 2 2 / 1 log (4x) + log₂        8 x2 = 8 – log₂2 _{4x + log} 2x2 – log28 = 8 – (log₂2 _{4 + log} 22x) + 2 log2x – 3 = 8 – 2 log₂2_{2 – log} 22x + 2 log2x – 11 = 0 163. 10 x 5 . 0 log x2 _{– 14 log} 16xx3 + 40 log4x x = 0 164. 6 – (1+4.94−2log33). log x 7 = logx7 6 –         + _4log4₉ 3 9 9 . 4 1 log₇x = _log1 _x 7 6 – _      + ₄ 33 log 2 3 36 . 4 1 log₇x = _log1 _x 7 6 –      + ₈ 3 36 . 4 1 165. log₃ (4.3x _{– 1) = 2x + 1} 32x + 1 _{= 4.3}x _{– 1} 32x _{. 3}1 _{= 4.3}x _{– 1} Let 3x _{= t} = 3t2 _{– 4t – 1 = 0 3}x ₌ 3 1 3t2 _{– 3t – t + 1 = 0} (3t – 1) (t – 1) = 0 3x _{= 1}

t = 3 1 , 1 x = 0 166. log₃ (3x _{– 6) = x – 1} 3x – 1 _{= 3}x _{– 6 3}x _{= 9} 3 3x = 3x _{– 6 x = 22} 3 t = t – 6 t = 3t – 18 – 2t = – 18 t = 9 167. log₃ (4x _{– 3) + log} 3 (4x – 1) = 1 Let 4x _{= t}

log₃(4x _{– 3) (4}x _{– 1) = 1 Not possible} 31 _{= (t – 3) (t – 1) 4}x _{= 4} 3 = t2 _{– 4t + 3 x = 1} 0 = t (t – 4)

t = 0, 4

168. log₃ (log_1/22_{x – 3 log}

1/2x + 5) = 2 x log2₂−1 – 3log₂−1x + 5 = 32 ⇒ – log₂2_{x + 3 log} 2x = 4 0 = log₂2_{x + 3 log} 2x + 4 0 = (log₂x _{– 1) (log} 2x + 4) ∴ log₂x _{= 4 log} 2x = 4 x = 2 x = 16 169. log₅ 2₁₀+x = x 1 2 5 log + log₅ 2₁₀+x – log₅ 1 x 2 + = 0 2 1 x 1 x 2 5 log + × + _{= 0} 20 x 2 x x 2 ₊ 2_{+ +} = 50 x2 _{+ 3x + 2 = 20} x2 _{+ 3x – 18 = 0} + 6x – 3x (x + 6) (x – 3) = 0 x = 3, – 6 Not satisfied 3 Ans. 170. 1 + 2 log_{x + 2}5 = log₅ x + 2 Let log₅x + 2 = t 5–1 _{= x + 2} x = ₅1 – 2 = – ₅9

1 + _log 2_{x 2} 5 + = log5 x + 2 log5x + 2 = 2 1 + 2_t = t 52 _{= x + 2} t + 2 = t2 _{x = 23} 0 = t2 _{– t – 2} 0 = (t – 2)–2 +1 _{(t + 1)} t = –1, 2 171. log₄24x ₌ 2log24 log₄24x _{= 4} 24x _{= 4}4 44x _{= (2}2₎4 24x _{= 8} 4x = 8 x = 2 172. log₂       4 x = ₁ 8 x log 15 2 − Let log2x = t

⇒ log₂x – log₂4 = _{log x} 15_{log 8 1} 2 2 − − ⇒ log₂x – 2 = _{log x}15_{3 1} 2 − − log₂x – 2 = _log15_{x 4} 2 − (t – 2) = _(t15₋₄₎ (t – 2) (t – 4) = 15 t2 _{– 2t – 4t + 8 = 15} t2 _{– 6t + 8 – 15 = 0} t2 _{– 6t – 7 = 0} t2 _{– 7t + t – 7 = 0} t (t – 7) + l(t – 7) = 0 (t + 1) (t – 7) = 0 t = –1, 7 log₂x = –1 or log₂x = 7 x = 2 1 x = 27 173. 2 2 2 ) x (log 2 x log ) x (log 2 1 − − =1 Let log x = t x log 2 x log ) x log 4 ( 2 1 2 10 10 2 10 − − = 1 log₁₀x = 0 1 – 8 log₁₀2_{x = log} 10x – 2 log102x x = 10° 0 = 6log₁₀2x + log₁₀x – 1

0 = (3 log₁₀x – 1) (2 log₁₀x + 1) = 0 ∴ 3 log₁₀x = 1 or 2 log₁₀x = – 1 log₁₀x = 3 1 log₁₀x = – 2 1 x = 3_{10 x =} 10 1 174. log₂ (4.3x _{– 6) – log} 2 (9x – 6) = 1 Let 3x = t log₂(4.t – 6) – log₂ (x2 _{– 6) = 1} log₂ 6 t 6 t 4 2₋ − = 1 4t – 6 = 2(t2 _{– 6)} 0 = 2t2 _{– 12 – 4t + 6} 0 = 2t2 _{– 4t – 6} 0 = t2 _{– 2t – 3} t2 _{– 3t + t – 3 = 0} ∴ t (t – 3) + 1(t – 3) = 0 (t + 1) (t – 3) = 0 ∴ t = 3, – 1 3x _{= 3} x = 1 3x _{= – 1 Rejected} ∴ Answer = 1 175. 2 1

log (5x – 4) + log x+1 = 2 + log 0₁₀₀.18

2 1

log₁₀ (5x – 4) +

2 1

log₁₀x + 1 = 2 + log₁₀18 – log₁₀100 ⇒ ₂1 log₁₀ (5x – 4) (x – 1) = log₁₀18 5x2 _{– 4x + 5x – 4 = 324} 5x2 _{+ x – 328 = 0} D = b2 _{– 4ac = 1 + 6560 = 6561} D = 6561 x ≡ – 10 82 , 8 ∴ ∴∴ ∴∴ Final Ans. 8 176. log₄ (2.4x – 2 _{– 1) + 4 = 2x} 2 2 log _       −1 4 21 . 2 ₂x _{+ 4 = 2x} 2 1 log₂        ₋ 16 16 2 . 2 2x + 4 = 2x log_{2 }       + ₋ 16 16 22x 1 + 4 = 4x

16 16 22x+1₋

177. log_x 5 + logx 5x – 2.25 = (logx 5 )2 log_x 5 + logx5 + logxx – 2.25 =

2 x50 log 2 11       x log 2 1 5 + log x 1 5 + 1 – 2.25 = 4log x 1 5 Let log5x = t t 2 1 + 1_t + 1 – 2.25 = ₄1_t = t 2 ) 25 . 2 ( t 2 t 2 2 1+ + − = t 4 1 2 + 4 + 4t – 4t (2.25) = 1 6 + 4 (t – 2.25 t) = 1 – 6 –1 – 1.75 t = 4 5 − = 7 5

178. log (log x) + log (log x4 _{– 3) = 0} log₁₀ [log x + 4 logx – 3] = 0 5 log x – 3 = 1 5 log₁₀x = 4 log₁₀x = 5 4 Let logx = t x = 104/5 ⇒ 5 log2_{x – 3 log ? = 0} log₁₀t + 4 log2_{t – 4 log 3}

179. log₃x – 2 log_1/3 x = 6 log₃x + 2log₃x = 6 log₃x + log₃x2 _{= 6} x3 _{= 3}6 x3 _{= (3}2₎3 x = 9 180. _log(2₅log_x−x₄₎ = 1 ⇒ 2 log x = log (5x – 4) log x2 _{= log (5x – 4)} x2 _{– 5x + 4 = 0} x2 _{– 4x – x + 4 = 0} x(x – 4) –1(x – 4) = 0 (x – 1) (x – 2) = 0 x = 1, 4 Answer = 4 181. 2 log₈2x + log₈x2 _{+ 1 – 2x =} log₈(2x)2 _{+ log} 8 x2 – 2x + 1 = ₃ 4

log₈4x2 _(x2 _{– 2x + 1) =} 3 4 4x2_(x2 _{– 2x + 1) = (2}3₎ 4x2_(x2 _{– 2x + 1) = 2}4 4x2 _(x2 _{– 2x + 1) = 16} x2 _[x2 _{+ 2x + 1] = 4} x4 _{+ 2x}3 _{+ x}2 _{– 4 = 0} 2 log₈2x + log₈(x2 _{+ 1 – 2x) =} 3 4

2[log₈2 + log₈x] + log₈ x2 _{– 2x + 1 =}

3 4

(x + 1) (x3 _{– 3x}2 _{+ 4x – 4)} (x – 2) (x2 _{– x + 2)}

Thus (x – 1) (x – 2) (x2 _{– x + 2) are its factor} since 1 can’t satisfies, 2 satisfies the answer ∴ Answer = 2 182. 6 1 log₂(x – 2) – 3 1 = log_{1/8 3x−5} 6 1 log₂ (x – 2) = 3 1 = log₂−3 3x−5 6 1 log₂ (x – 2) – ₃1 = – ₃1 log₂(3x – 5)1/2 6 1 log₂ (x – 2)1/6 _{. (3x – 5)}1/6 ₌ 3 1 = log₂ (x – 2)1/6 _{. (3x – 5)}1/6 ₌ 3 1 ((x – 2)(3x – 5))1/6 _{= (2)}1/3 × 6 if 9 multiplied power by, get 3x2 _{– 6x –5x + 10 = 4} 3x2 _{– 11x + 6 = 0} 3x (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 _{= 0} log₃(x – 2)2 _{+ log} 3 (x – 4)2 = 0 2 log₃(x – 2) + 2 log₃ (x – 4) = 0 2[log₃(x – 2) (x – 4)] = 0 x2 _{– 2x – 4x + 8 = 3° D = b}2 _{– 4ax} x2 _{– 6x + 8 – 1 = 0 36 – 28 = 8} x2 _{– 6x + 7 = 0 D = 2 2}

∴ x = 2 2 2 6± = 2 2 ) 2 3 ( ± = 3 + 2, 3 – 2 184. log (2x2).log₄16x

2 = log4x3 let log2x = t

) x log 16 )(log x log 2 (log 2 _{4 4} 2 2 + + = 3 log4x ⇒ (1+2log2x)_ +2 1₂log2x_ = 2 3 log₂x = 2 2 t 2 ) t 2 1 ( _             + + ₌ t 2 2 3       (1 + 2t)       + 2 t 4 = 4 t 9 2 2 (4 + 8t + t + 2t2_{) = 9t}2 0 = 9t2 _{– 4t}2 _{– 18t – 8} 0 = 5t2 _{– 18t – 8} 0 = 5t2 _{– 20t + 2t – 8} 0 = 5t (t – 4) + 2(t – 4) 0 = (5t + 2) (t – 4) ∴ t = 4, – 5 2 ∴ log₂x = 4 log₂x = – 5 2 x = 24 _{x = (2)}–2/5 _{can’t satisfied me} x = 16

185. 3₃log_logx_x+₋19₁ = 2 logx + 1 Let log₁₀x = t

1 t 3 19 t 3 − + = 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 – 20} 0 = 3t2 _{– t – 10} 0 = 3t2 _{– 6t + 5t – 10} 0 = (3t + 5) (9t – 2) 186. 40 x log ) 1 1 x log( 3 ₋ + + = 3

log ( x+1+1) = 3 _log3x − 40_ log x+1 + 1 = logx – 120 120 = logx – log x+1 – 1 187. log₃2_{6 – log} 32 2 = (log2x – 2) log3121 log₃2 _{2 + log}

323 – log322 = (log2x – 2) log312 (log₃x)2 _{= (log} 102x – 2) (log33 + log34) 188. 1 – 2 1 log (2x – 1) = 2 1 log (x – 9) 1 = 2 1 log (x – 9) + 2 1 log (2x – 1) 1 = 2 1 [log (x – 9) (2x – 1)] 2 = log₁₀ 2x2 _{– 19x + 9} 102 _{= 2x}2 _{– 19x + 9} 0 = 2x2 _{– 19x – 91} 0 = 2x2 _{– 26x + 7x – 91} 0 = 2x (x – 13) + 7(x – 13) 0 = (2x + 7) (x – 13) x = – 2 7 , 13 x = 13 Ans.

189. log_log8x₋+_log(7−_xlog₋₅2₎ = – 1

5 x 8 10 2 7 x 10 log log − + = – 1 log₁₀ 2 7 x+ = ? – log_{10 x}8₋₅ 190. log(3x2 _{+ 7) = log (3x – 2) = 1} log (3x2 _{+ 7) = 1 log (3x – 2) = 1} 101 _{= 3x}2 _{+ 7 10}1 _{= 3x – 2} 3 = 3x2 _{12 = 3x} x = 1 x = 4 log (3x2 _{+ 7) – log (3x – 2) = 0} log 2 x 3 7 x 3 2 − + = 0 3x2 _{+ 7 = 3x – 2} 3x2 _{– 3x + 9 = 0} x2 _{– x + 3 = 0} 191. 3 1 log(x2 _{– 16x + 20) – log 3} 7 = ₃1log (8 – x)

3 1 [log₁₀x2 _{– 16x + 20 – log} 107] = ₃ 1 log₁₀8 – x log₁₀ 7 20 x 16 x2− + = log₁₀8 – x x2 _{– 16x + 20 = 56 – 7x} x2 _{– 16x + 20 – 56 + 7x = 0} x2 _{– 9x – 36 = 0} – 12 + 3 x(x – 12) + 3(x – 12) = 0 (x + 3) (x – 12) = 0 ∴ x = –3, 12 makes the x < 1 rejected x = –3 Ans. 192. log₆3x + 3 _{– log} 6 3x – 2 = x log_{6 x 2} 3 x 3 2 − + = x 2 x 3 x 3 2 − + = 6x 2 x 3 x 3 2 − + = 2x _{. 3}x 2 x x x 3 3 3 2 = 8 2x × ₃x 9 = 2x _.3x 193.       + x 2 1

1 _{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 log} 2(2x + 3 – 6) log₂ (22x _{+ 1) = x + log} 2 _      − 6 8 2x 198. log₃ (9x _{+ 9) = x + log} 3 (28 – 2 . 3x)

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

4 log x – 2 = 1 log x =

4 3

x = 103/4

200. log ₅4x _{– 6 = –} 5 log 2x_{– 1 = 2} 5 log 2 2 6 4 x x − − = 2 Let 2x _{= t} 2 log₅ 2 t 6 t2 − − = 2 2x _{= 4} log₅ 2 t 6 t2 − − = 1 2x _{= 4 x = 6} 2 t 6 t2 − − = 5a1 _{∴ x = 2} t2 _{– 6 = 5t – 10} t2 _{– 5t + 4 = 0} t(t – 1) – 4(t – 1) = 0 t = 4, 1 203. log _       − _x4 x 2 2 3 _{= 2 +} 4 1 log 16 – 2 4 log x 204. x log x log2 5 1_ −      = 125 1 . 5logx – 1 205. 3logx 2 x log 3 2 x − = 100310 206. log₂ (25x + 3 _{– 1) = 2 + log} 2 (5 x + 3 + 1) 210. log₂(2x2_{) . log} 2 (16x) = ₂ 9 log₂2_x (log₂2 _{+ log} 2x2) (4 + log24) = ₂ 9 log₂2 _x (1 + 2 log₂x) (4 + log₂x) = 2 9 log₂2_x

⇒ 2 [4 + 8 log₂x + log₂x + 2log₂2_{x] = 9 log} 22x ⇒ 8 + 16 log₂ x + 2 log₂x + 4 log₂2_{x = 9 log}

22 x 0 = 5 log₂2_{x – 18 log} 2x – 8 Let log2x = t 0 = 5t2 _{– 18t – 8t} D = b2 _{– 4ac} = 324 + 160 = 484 D = 22 t = 10 22 18± = 10 40 = 4 ∴ log₂x = 4 log₂x = – ₅2 x = 24 _{x = (2)}– 2/5

16 = 2 211. log₄4 +       + x 2 1 1 _{log3 = log }     3_{3 + 27} since radical should be defined log₄4 + 24 1 1 3 log + = log _     x_{3 + 27}

log (4.3 . 31/2x_{) = log (3}1/x _{+ 27) Let 3}1/x _{= t}

12.t = t2 _{+ 27 3}1/x _{= t} 0 = t2 _{– 12t + 27} 0 = t2 _{– 9t – 3t + 27} 0 = (t – 9) (t – 3) ∴ t = 9, 3 x = 1, 2 1

212. log(x3 _{+ 27) – log (x}2 _{+ 6x + 9) = log} 3 ₇ log (x + 3) (x2 _{– 3x + 9) – log} 2 2 / 1 2 ) 3 x ( ) 9 x 6 x ( + + + = log 7 log₁₀         + + − + ) 3 x ( ) 9 x 3 x )( 3 x ( 2 = log₁₀7 x2 _{– 3x + 9 – 7 = 0} x2 _{– 3x + 2 = 0} x2 _{– 2x – x + 2 = 0} x(x – 2) – 1(x – 2) = 0 (x – 1) (x – 2) = 0 x = 1, 2 213. 5log x _{= 50 – x}log 5

5log x _{+ x}log 5 _{= 50 Let log} 10x = t 5log x _{+ 5}log x _{= 50 x = 10}2 _{= 100} 5t _{+ 5}t _{= 50} 2.5t _{= 50} 5t _{= 25} t = 2 215. logx 1 x log 3 x − = 310 x log 1 x log 3 2 x − = 310 log logx 1 x log 3 2 x − = log 3₁₀ log x . 3log_logx_x 1

2 ₋ = 3 1 log 10 3 log2_{x – 1 =} 3 1

3 log2_{x =} 3 4 log₁₀2_{x =} 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) log} 2(x – 3) = 2(x – 10) log₂(x – 3) = 2 log₂(x – 3) = – 2 x – 3 = 29 _{(x – 3) = 2}–2 x = 7 (x – 3) = 4 1

⇒

x = 4 1 + 3 = 4 13

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

2 1

log₂ log₂x + log₂

2 1 log₂x = 2 2x _{= 4} 2 1 log₂t + log ₂ 2 t = 2 x = 24 log₂ t1/2 _{+ log} 2t – log22 = 2 log₂t1/2 _{+ log} 2t – log22 = 2 log₂t1/2 _{+ log} 2t = 3 2 1 log₂t + log₂t = 3 log₂t + 2 log₂t = 6 3 log₂ t = 6 t = 22 _{= 4} 218. (6x – 5) |ln (2x + 2.3)| = 8 ln (2x + 2.3) (6x – 5) = _|8_nn₍(₂2_xx₊+₂2_..₃3₎)_| l l Ist 6x – 5 = – 8 6x = 13 x = 13₆ = – 2 1 It is not in answer IInd 6x – 5 = 8 The right answer

6x = 13 = 13₆ , −₂₀13 x = 13₆

219. log₉9x8−log₃3x = log₃x3

= 2 3 3x.(1 log x) log 2 8 1 _       + + _{= (3 log} 3x)2 = (1 + 4 log₃x) (1 + log₃x) = 9 log₃2 _x (1 + 4t) (1 + t) = 9t2 0 = 9t2 _{– 4t}2 _{– 5t – 1} 0 = 5t2 _{– 5t – 1} D = b2 _{– 4ac = 25 + 20 = 45} t = 10 5 3 5± log₃x = t log₃ x = 10 5 3 5± x = 3 10 5 3 5±

220. log2 _{(100x) – log}2 _{10x + log}2_{x = 6} log₁₀2 ₁₀2 _{+ log}

102x – [log21010+log102 x] + log102x = 6 4 + log 1 log2 x 10 2 10− − = 6 2 10 log x = 3 log₁₀x = 3 221. 9log1/3(x+1)= ₅log1/5(2x2+1) ) 1 x ( log 2 ₃1 3 − + = log₅1(2x2 1) 5 − + 2 3(x 1) log 2 3− + − = (2x2 + 1)–1 (x + 1)–2 _{= (2x}2 _{+ 1)}–1 2 ) 1 x ( 1 + = 2x 1 1 2₊ 2x2 _{+ 1 = x}2 _{+ 1 + 2x} 2x2 _{– x}2 _{– 2x = 0} x2 _{– 2x = 0} x(x – 2) = 0 x = 0, 2 223. 2 log₂       − − 1 x 7 x + log₂ __xx₊−₁1_ = 1 log₂ x_(x 49_1)(x14₁₎x 2 − − − + × 1 x 1 x + − = 1 ⇒ 1 x x 14 490 x 2 2 − − + = 2 ⇒ x2 _{+ 49 – 14x = 2x}2 _{– 2} ⇒ x2 _{+ 14x – 2 – 49 = 0} x2 _{+ 14x – 51 = 0} x2 _{+ 17x – 3x – 51 = 0}

x(x + 17) – 3(x + 17) = 0

∴ x = – 17, 3→can’t satisfy the equation. ∴ x = – 17 only 225. 2 1 log (1 + x) + 2 3 log (1 – x) = 2 1 log (1 – x2₎ = 2 1 [log (1 + x) + log (1 – x)3_{] =} 2 1 log (1 – x2₎ log _(1−x1₎2₍x_1−x) + = log (1 – x2₎ ⇒ log _(1−x1₎2₍x_1−x) + – log(1 – x2_{) = 0} 2 3 10 x 1 ) x 1 ( log x 1 − − + = 0 3 ) x 1 ( ) x 1 ( − + × _(1+x)(1_1−x) = 1 log_{10 (1 x)}4 1 − = 0 log₁₀ (1 – x)–4 _{= 0} – 4 log₁₀ (1 – x) = 0 100 _{= 1 – x} x = 0 226. log(_log(35_{5 x}x₎) 3 − − = 3 log (35 – x3_{) = 3 log (5 – x)} log (35 – x3_{) = log(5 – x)}3 35 – x3 _{= 125 – x}3 _{– 15x (5 – x)} 35 – 125 = – 75x + 15x2 0 = 15x2 _{– 75 x + 90} 0 = x2 _{– 5x + 6} 0 = x2 _{– 2x – 3x + 6} ∴ x = 2, 3 227. log_x2 – log₄x + 7₆ = 0 x log 1 2 – 2 1 log₂x + ₆7 = 0 6 – 3 log₂2 _{x ± 7 log} 2x = 0 3 log₂2 _{x – 9 log} 2x + 2 log2x – 6 = 0 (3 log₂x + 2) (log₂x – 3) = 0 3 log₂x = – 2 log₂x = 3 log₂x = – ₃2 x = 23 x = 2 – 2/3 _{x = 8}

229. log₂x – 0.5 = log_{2 x} x log₂ = 2 1 log₂x + 2 1 x log₂ = 2 1 [log₂x + 1] 4 log₂x = log₂2_{x + 1 + 2 log}

2x 0 = log₂2 _{x – 2 log}

2x + 1

0 = log₂x (log₂x – 1) – 1(log₂x – 1) 0 = log₂x – 1 1 = log₂x x = 2 230. log_{1/3 }       −     × 1 2 1 2 x = log_{1/3 }       −     4 2 1 2x 2 x 2 1     – 1 = x 2 2 1     – 4 Let x 2 1       = t 2 x 2 1_      –       2 1 . x 2 1_      = – 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 2 1_      = t x 2 1_      = – 1, 3 –1 not defined log x 2 1       = log 3 log 2–x _{= log 3} – x log 2 = log 3 – x = _loglog₂3 x = – log₂3 Ans. 231. 2 1 log₆(x – 2) + 2 1 log₆(x – 11) = 1 2 1 log₆ (x – 2) (x – 11) = 1 log₆ x2 _{– 13x + 22 = 2} x2 _{– 13x + 22 = 6}2 x2 _{– 13x + 22 – 36 = 0} x2 _{– 13x – 14 = 0} x2 _{– 14x + x – 14 = 0} x(x – 14) + 1(x – 14) = 0

(x + 1)(x – 14) = 0 x = 14 Ans. –1 is not defined

233. log₇ 2 + log x = ₇2 log₇−1(3)1/2 log₇2 + 2 1 log₇x = – 2 1 log₇3 log₇2 + 2 1 log₇3 + 2 1 log₇x = 0 2 log₇2 + log₇ 3 + log₇x = 0 log₇ 12 = – log₇x

x = – 12

237. log_3x

x 3

+ log₃2_{x = 1 Let log} 3x = t x 3 log x 3 log 3 3 + log₃2 _{x = 1 log} 3x = 0 x log 3 log x log 3 log 3 3 3 3 + − + log 32_{x = 1 log} 3x = 1 t 1 t 1 + − + t2 _{= 1 x = 1} 1 – t + t2 _{(1 + t) = 1 + t x = 3}–2 t3 _{+ t}2 _{– 2t = 0} t(t2 _{+ t – 2) = 0} t2 _{+ t – 2 = 0} t(t + 2) – 1(t + 2) = 0 t = 1, 2 Ans. 1, 3, 3–2 246. log₄(x + 12) log_x2 = 1 2 1 log _logx 12_x 2 + = 1 log₂ (x + 12) = 2 log₂x log₂ x + 12 = log₂x2 x + 12 = x2 0 = x2 _{– x – 12} 0 = x2 _{– 4x + 3x – 12} 0 = x(x – 4) + 3(x – 4) x = 4, 3 not satisfied ∴ 24 Ans.

247. log₁₆x + log₄x + log₂x = 7

4 1 log₂x + 2 1 log₂x + log₂x = 7 4 x log 4 x log 2 x log2 + + 2 _{= 7} 7 log₂x = 28 log₂x = 4

x = 16 248. log 10 + 3 1 log (32 x ₊271)_{= 2} 3 1 log (32 x ₊271) _{= 1} log₁₀ (32 x +271) = 3 x 2 3 + 271 = 103 x 2 3 = 1000 – 271 x 2 3 = 279 x 2 3 = 36 ⇒ 2 x = 6 x = 3 ⇒ _{( x)}2 _{= 9 ⇒ x = 9} 250. log₅ (3x _{+ 10) + 7.1}0 _{= log} 5 (9x + 156)

Detective Question for me and how can I solve

251. (log₄x – 2) log₄x =

2 3

(log₄x – 1)

2(log₄x – 2) log₄x = 3 log₄x – 3 Let log₄x = t (2t – 4) t = 3t – 3 2t2 _{– 4t – 3t + 3 = 0} 2t2 _{– 7t + 3 = 0} 2t2 _{– 6t + t + 3 = 0} 2t (t – 3) – 1(t – 3) = 0 t = 3, 2 1 ∴ log₄x = 3, 2 1 x = 43 _{, (4)}1/2 x = 64, 2

252. log₅x + log₂₅x = log_1/5 3 log₅x + 2 1 log₅x = – 2 1 log₅ 3 2 log₅x + log₅x = – log 53 – 3 log₅x = log₅3 log₅x–3 _{= log} 5 3 (x–3₎1/3 _{= (3)}– 1/3 x = 3₃ 1 254. log₂ (9 – 2x_{) = 3 – x} 23 – x _{= 9 – 2}x _{Let 2}x _{= t}

x 3 2 2 = 9 – 2x ₂x _{= 1} t 8 = 9 – t 2x _{= 8} 8 = 9t – t2 _{x = 3} t2 _{– 9t + 8 = 0} t2 _{– 8t – t + 8 = 0} t (t – 8) – 1 (t – 8) = 0 (t – 1) (t – 8) = 0 t = 1, 8 x = 0, x = 3 Ans.

255. 2(log 2 – 1) + log (5 x ₊1) _{+ log (5}1− x ₊₅₎ 2(log 2 – 1) + log (5 x +1) + log _

     + 5 5 5 x

256. log_x3 + log₃x = log _x3 + log₃ x +

2 1 x log 1 3 + log3x = log x 2 3 + 2 1 log₃x + 2 1 t 1 + t = 2_t + 2 t + 2 1 t t 1+ 2 = t 2 t t 2+ 2+ 2 + 2t2 _{= 2 + t}2 _{+ t} t2 _{– t = 0 Q log} 3x = 0 t = 0, 1 x = 30 _{= 1} log₃x = 1 ⇒ x = 3

259. log_x 125x . log2₂₅x= 1 (log₅2 x)2

x log x 125 log 5 5 . log₅2_{2x = 1} 2 5x log 2 1       x log x log 5 log 5 5 3 5 + . 2 1 log2 5x = 1 t t 3+ . 4 t2 = 1 t 4 t t 3 2+ 3 = 1 t(3t + t2_{) = 4t} t2 _{+ 3t – 4 = 0} t2 _{+ 4t – t – 4 = 0} t(t + 4) – 1(t + 4) = 0 (t – 1) (t + 4) = 0 t = 1, – 4 ∴ log₅x = 1, –4 x = 5, 5–4