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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.

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    $\begingroup$ Just group terms containing $x$ $\endgroup$ – lab bhattacharjee Jan 2 '14 at 16:37
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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.

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Your almost done, you just need to collect the $x$-terms: $$2x\log(7)+x\log(4)-2\log(4)-x\log(11)=0$$ $$x(2\log(7)+\log(4)-\log(11))=2\log4$$ $$x=\frac{2\log(4)}{2\log(7)+\log(4)-\log(11)}$$

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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 $$ x(2\log(7) + \log(4) - \log(11)) = 2\log(4) \\ x = \frac{2\log(4)}{2\log(7) + \log(4) - \log(11)}. $$

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