Juan Leonard

2022-10-21

Logarithms equation with tricky transformations
${8}^{x-2}×{5}^{x+2}=1$
This one according to wolfram alpha it has nice solution
$x=\frac{2\left(\mathrm{log}\left(8\right)-\mathrm{log}\left(5\right)\right)}{\mathrm{log}\left(8\right)+\mathrm{log}\left(5\right)}$
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.

Do you have a similar question?

giosgi5

Expert

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

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