Solve $2^x\cdot 6^{x-2}=5^{2x}\cdot 7^{1-x}$

Question

I have to solve this equation, $2^x\cdot 6^{x-2}=5^{2x}\cdot 7^{1-x}$. Now, I started by taking logs on both sides which gives me this funny looking equation

$x\log{2}+(x-2)\log(2\cdot3)=2x\log(\frac{10}{2})+(1-x)\log{7}$

I have been stuck on this step for a while now and can't see how I can go further from here. Is there a way out?

Answer 1

It's $$x\ln2+(x-2)\ln6=2x\ln5+(1-x)\ln7,$$ which gives $$x=\frac{2\ln6+\ln7}{\ln2+\ln6-2\ln5+\ln7}$$ or $$x=\log_{\frac{84}{25}}252$$

Answer 2

$$\frac{2^\color{red}{x}\times6^\color{red}{x}}{6^2}=\frac{(5^2)^\color{red}{x}\times7^1}{7^\color{red}{x}}$$ $$\frac{2^\color{red}{x}\times6^\color{red}{x}\times7^\color{red}{x}}{(5^2)^\color{red}{x}}=7^1\times6^2$$ $$\Big(\frac{2\times6\times7}{25}\Big)^\color{red}{x}=7^1\times6^2$$ so $$\color{red}{x}=\frac{\ln\Big(7^1\times6^2\Big)}{\ln\Big(\frac{2\times6\times7}{25}\Big)}$$

Answer 3

$$\begin{matrix} 2^x.6^{x-2} & = & 5^{2x}.7^{1-x} \\ \left(\dfrac{2*6*7}{5^2}\right)^x & = & 6^2 * 7 \\ \left(\dfrac{84}{25}\right)^x & = & 252 \\ x & = & \dfrac{\ln\left(252\right)}{\ln\left(\dfrac{84}{25}\right)} \end{matrix}$$