# Solving $7^{2x}\cdot4^{x-2}=11^x$

I am trying to solve this equation but I am stuck little bit right now. This is how I did it: $$7^{2x}\cdot4^{x-2}=11^x\\ \text{log both sides}\\ \log(7^{2x}\cdot4^{x-2})=\log(11^x)\\ \log(7^{2x}) + \log(4^{x-2})=\log(11^x)\\ 2x\log(7) + (x-2)\log (4)=x\log (11)\\ \text{I got this far}\\$$

Can somebody give me idea, am I on right track, and if I am, how can I solve it further. Thanks.

• Just group terms containing $x$ – lab bhattacharjee Jan 2 '14 at 16:37

You've done most of it. Next you get $$x(2\log7+\log4) -2\log4 = x\log11,$$ and so $$x(2\log7+\log4-\log11) = 2\log4$$ etc.
Your almost done, you just need to collect the $x$-terms: $$2x\log(7)+x\log(4)-2\log(4)-x\log(11)=0$$ $$x(2\log(7)+\log(4)-\log(11))=2\log4$$ $$x=\frac{2\log(4)}{2\log(7)+\log(4)-\log(11)}$$
Yes, you are on the right track. Now just collect the terms with $x$ in it on one side and the constant on the other side. You should get $$x(2\log(7) + \log(4) - \log(11)) = 2\log(4) \\ x = \frac{2\log(4)}{2\log(7) + \log(4) - \log(11)}.$$