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$$8^{x-2}\times5^{x+2}=1$$

This one according to wolfram alpha it has nice solution $$x = \frac{2 (\log(8)-\log(5))}{\log(8)+\log(5)}$$

I see one could guess this solution and just assume left side is increasing function and be done, but I want to see some transformations which could bring me to this solution and I'm stuck.

BLAZE
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Marcin Piekarniak
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3 Answers3

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$$ 8^{x-2}\times5^{x+2}= 8^{-2} \cdot 5^2 \cdot 8^x\cdot 5^x = \frac{5^2}{8^2} \cdot 40^x. $$ That is equal to $1$ precisely if $$ 40^x = \frac{64}{25} $$ and that holds only if $$ x = \log_{40}\frac{64}{25}. $$

Michael Hardy
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$$\log(a^n.b^m)=n\log a+m\log b$$ take the $\log$ $$(x-2)\log 8+(x+2)\log 5=0$$

E.H.E
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$$ a^nb^n = (ab)^n\\ a^{n+m} = a^na^m\\ \log(1) = 0 $$ can you use these rules to determine the value of $x$?

Chinny84
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