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