rigliztetbf

2022-06-29

Solving ${7}^{2x}\cdot {4}^{x-2}={11}^{x}$
I am trying to solve this equation but I am stuck little bit right now. This is how I did it:
${7}^{2x}\cdot {4}^{x-2}={11}^{x}\phantom{\rule{0ex}{0ex}}\text{log both sides}\phantom{\rule{0ex}{0ex}}\mathrm{log}\left({7}^{2x}\cdot {4}^{x-2}\right)=\mathrm{log}\left({11}^{x}\right)\phantom{\rule{0ex}{0ex}}\mathrm{log}\left({7}^{2x}\right)+\mathrm{log}\left({4}^{x-2}\right)=\mathrm{log}\left({11}^{x}\right)\phantom{\rule{0ex}{0ex}}2x\mathrm{log}\left(7\right)+\left(x-2\right)\mathrm{log}\left(4\right)=x\mathrm{log}\left(11\right)\phantom{\rule{0ex}{0ex}}\text{I got this far}\phantom{\rule{0ex}{0ex}}$
Can somebody give me idea, am I on right track, and if I am, how can I solve it further. Thanks.

Jerome Page

Expert

You've done most of it. Next you get
$x\left(2\mathrm{log}7+\mathrm{log}4\right)-2\mathrm{log}4=x\mathrm{log}11,$
and so
$x\left(2\mathrm{log}7+\mathrm{log}4-\mathrm{log}11\right)=2\mathrm{log}4$
etc.

Garrett Black

Expert

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

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