Question

solve 3^(2x) = 7^(x-1) −7.74293 7.74293 −1 1

Answer 1

Hello,

3^(2x)=7^(x-1)
==>2x ln 3 = (x-1)ln 7
==>x(2ln 3 -ln 7)=-ln 7
==>x=-ln 7 /(2 ln 3-ln 7)
==>x=-7,7429304889740634853601918674823...

Answer A

Answer 2

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.