Q116 / 200 · TS EAMCET 2002 · Mathematics

If $\log 2=a, \log 3=b, \log 7=c$ and $6^x=7^{x+4}$ then $x$ is equal to

  • A
    $\frac{4 b}{c+a-b}$
  • Correct answer
    B
    $\frac{4 c}{a+b-c}$
  • C
    $\frac{4 c}{c-a-b}$
  • D
    $\frac{4 a}{a+b-c}$

Solution

We have, $\begin{array}{cc} & 6^x=7^{x+4} \\ \Rightarrow & x \log 6=(x+4) \log 7 \\ \Rightarrow & x(\log 2+\log 3)=x \log 7+4 \log 7 \\ \Rightarrow & x(\log 2+\log 3-\log 7)=4 \log 7 \\ \Rightarrow & x=\frac{4 \log 7}{\log 2+\log 3-\log 7}=\frac{4 c}{a+b-c} \end{array}$

Hence, the correct answer is option: 'B'.

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