rigliztetbf

Answered

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:

${7}^{2x}\cdot {4}^{x-2}={11}^{x}\phantom{\rule{0ex}{0ex}}\text{log both sides}\phantom{\rule{0ex}{0ex}}\mathrm{log}({7}^{2x}\cdot {4}^{x-2})=\mathrm{log}({11}^{x})\phantom{\rule{0ex}{0ex}}\mathrm{log}({7}^{2x})+\mathrm{log}({4}^{x-2})=\mathrm{log}({11}^{x})\phantom{\rule{0ex}{0ex}}2x\mathrm{log}(7)+(x-2)\mathrm{log}(4)=x\mathrm{log}(11)\phantom{\rule{0ex}{0ex}}\text{I got this far}\phantom{\rule{0ex}{0ex}}$

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

I am trying to solve this equation but I am stuck little bit right now. This is how I did it:

${7}^{2x}\cdot {4}^{x-2}={11}^{x}\phantom{\rule{0ex}{0ex}}\text{log both sides}\phantom{\rule{0ex}{0ex}}\mathrm{log}({7}^{2x}\cdot {4}^{x-2})=\mathrm{log}({11}^{x})\phantom{\rule{0ex}{0ex}}\mathrm{log}({7}^{2x})+\mathrm{log}({4}^{x-2})=\mathrm{log}({11}^{x})\phantom{\rule{0ex}{0ex}}2x\mathrm{log}(7)+(x-2)\mathrm{log}(4)=x\mathrm{log}(11)\phantom{\rule{0ex}{0ex}}\text{I got this far}\phantom{\rule{0ex}{0ex}}$

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

Answer & Explanation

Jerome Page

Expert

2022-06-30Added 16 answers

You've done most of it. Next you get

$x(2\mathrm{log}7+\mathrm{log}4)-2\mathrm{log}4=x\mathrm{log}11,$

and so

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

etc.

$x(2\mathrm{log}7+\mathrm{log}4)-2\mathrm{log}4=x\mathrm{log}11,$

and so

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

etc.

Garrett Black

Expert

2022-07-01Added 5 answers

Your almost done, you just need to collect the $x$-terms:

$2x\mathrm{log}(7)+x\mathrm{log}(4)-2\mathrm{log}(4)-x\mathrm{log}(11)=0$

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

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

$2x\mathrm{log}(7)+x\mathrm{log}(4)-2\mathrm{log}(4)-x\mathrm{log}(11)=0$

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

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

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