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Mathematics 7 Online
OpenStudy (anonymous):

5^(x+2)= 2^(3-x)

OpenStudy (anonymous):

What do you want to do with the problem? Factor Simplify Solve for x ??

OpenStudy (anonymous):

Solve for X

OpenStudy (solomonzelman):

Hit both sides with log.

OpenStudy (solomonzelman):

\(\Huge\color{blue}{ \sf 5^{(x+2)}= 2^{(3-x)} }\) \(\Large\color{blue}{ \sf (x+2)Log5= (3-x)Log2 }\)

OpenStudy (anonymous):

I logged both with the same log

OpenStudy (solomonzelman):

\(\Large\color{blue}{ \sf \frac{(x+2)}{(3-x)} = \frac{Log5}{Log2}=Log_25 }\)

OpenStudy (anonymous):

Ok I understand the first two parts of your last post but not your final answer.

OpenStudy (anonymous):

I am solving for X so does x = \[\log_{2} 5\]?

OpenStudy (kc_kennylau):

Nope

OpenStudy (kc_kennylau):

Don't divide

OpenStudy (kc_kennylau):

Expand (x+2)Log5=(3−x)Log2

OpenStudy (kc_kennylau):

\[x\log5+2\log5=3\log2-x\log2\]

OpenStudy (kc_kennylau):

\[x(\log5+\log2)=3\log2-2\log5\]

OpenStudy (kc_kennylau):

\[x=\frac{3\log2-2\log5}{\log5+\log2}=\cdots\]

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