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If $\log 2=a, \log 3=b, \log 7=c$ and $6^x=7^{x+4}$ then $x$ is equal to
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Verified Answer
The correct answer is:
$\frac{4 c}{a+b-c}$
We have,
$\begin{array}{cc}
& 6^x=7^{x+4} \\
\Rightarrow & x \log 6=(x+4) \log 7 \\
\Rightarrow & x(\log 2+\log 3)=x \log 7+4 \log 7 \\
\Rightarrow & x(\log 2+\log 3-\log 7)=4 \log 7 \\
\Rightarrow & x=\frac{4 \log 7}{\log 2+\log 3-\log 7}=\frac{4 c}{a+b-c}
\end{array}$
$\begin{array}{cc}
& 6^x=7^{x+4} \\
\Rightarrow & x \log 6=(x+4) \log 7 \\
\Rightarrow & x(\log 2+\log 3)=x \log 7+4 \log 7 \\
\Rightarrow & x(\log 2+\log 3-\log 7)=4 \log 7 \\
\Rightarrow & x=\frac{4 \log 7}{\log 2+\log 3-\log 7}=\frac{4 c}{a+b-c}
\end{array}$
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