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7^x-2 = 5^3-x
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\[7^{x-2} = 5^{3-x} \LARGE\] ?
yes, that is it. 7^(x-2) = 5^(3-x)
I don't think you can solve this, try it your own
it has to do with logs
Yes, but you can't find a real solution
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i can
\[\begin{align}7^{x-2}&=5^{3-x}\\ \frac{7^x}{7^2}&=\frac{5^3}{5^x}\\ 5^x\cdot7^x&=5^3\cdot7^2\\ (5\times7)^x&=5^3\cdot7^2\\ \log_{(5\times7)}(5\times7)^x&=\log_{(5\times7)}(5^3\cdot7^2)\\ x&=\frac{\log(5^3\cdot7^2)}{\log{(5\times7)}} \end{align}\]
or ln both sides, lol
damme wolf gives 'no solution', I guess im just being lazy :P
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