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Question

Mathematics Question on Exponential and Logarithmic Functions

If 3x=4x−13^x = 4^{ x - 1} then x is equal to

A

2 log3 22log32−1 \frac{ 2 \, log_3 \, 2}{ 2 log_3 2 - 1}

B

22log23 \frac{ 2}{ 2 log_2 3 }

C

11−log43\frac{ 1}{ 1 - log_4 3 }

D

2log2 32log23−1 \frac{ 2 log_2 \, 3} { 2 log_2 3 - 1}

Answer

11−log43\frac{ 1}{ 1 - log_4 3 }

Explanation

Solution

3x=4x−13^x = 4^{ x - 1} Taking log3log_3 on both sides, we get ⇒x log33=(x−1)log34⇒x=2log32.x−log34\Rightarrow x \, log_3 3 = (x - 1) log_3^4 \Rightarrow x = 2 log_3 2 . x - log_3 4 ⇒x(1−2 log3 2)=−2 log32⇒x=2log322log32−1\Rightarrow x ( 1 - 2 \, log_3 \, 2) = - 2 \, log_3 2 \Rightarrow x = \frac{ 2 log_3 2 }{ 2 log_3 2 - 1} x = 11−12log32=11−1log34=11−log43=22−log23\frac{1}{ 1 - \frac{1}{ 2 log_3 2 } } = \frac{1}{ 1 - \frac{1}{ log_3 4 } } = \frac{1}{ 1 - log_4 3 } = \frac{2}{ 2 - log_2 3} = 11−12log23=11−log43 \frac{1}{ 1 - \frac{1}{2} log_2 3 } = \frac{1}{ 1 - log_4 3 }