Answer:
Option (a) is correct.
\(x=-7.74293\)
Step-by-step explanation:
Given : \(\:3^{\left(2x\right)}\:=\:7^{\left(x-1\right)}\)
We have to solve the given expression \(\:3^{\left(2x\right)}\:=\:7^{\left(x-1\right)}\)
Consider the given expression, \(\:3^{\left(2x\right)}\:=\:7^{\left(x-1\right)}\)
Using, \(\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)\) , we have,
\(\ln \left(3^{2x}\right)=\ln \left(7^{x-1}\right)\)
Apply log rule,
\(\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)\)
We have,
\(\ln \left(3^{2x}\right)=2x\ln \left(3\right),\:\space\ln \left(7^{x-1}\right)=\left(x-1\right)\ln \left(7\right)\)
substitute, we get,
\(2x\ln \left(3\right)=\left(x-1\right)\ln \left(7\right)\)
\(\mathrm{Expand\:}\left(x-1\right)\ln \left(7\right):\quad \ln \left(7\right)x-\ln \left(7\right)\)
Subtract \(\ln \left(7\right)x\) from both side, we have,
\(2x\ln \left(3\right)-\ln \left(7\right)x=\ln \left(7\right)x-\ln \left(7\right)-\ln \left(7\right)x\)
Simplify, we have,
\(2x\ln \left(3\right)-\ln \left(7\right)x=-\ln \left(7\right)\)
Taking x common from both term , we have,
\(x\left(2\ln \left(3\right)-\ln \left(7\right)\right)\)
Divide both side by \(2\ln \left(3\right)-\ln \left(7\right)\) , we have,
\(\frac{\left(2\ln \left(3\right)-\ln \left(7\right)\right)x}{2\ln \left(3\right)-\ln \left(7\right)}=\frac{-\ln \left(7\right)}{2\ln \left(3\right)-\ln \left(7\right)}\)
On simplify , we get,
\(x=-\frac{\ln \left(7\right)}{\ln \left(\frac{9}{7}\right)}\)
Thus, \(x=-7.74293.....\)
Thus, option (a) is correct.
