Juan Leonard

2022-10-21

Logarithms equation with tricky transformations

$${8}^{x-2}\times {5}^{x+2}=1$$

This one according to wolfram alpha it has nice solution

$$x=\frac{2(\mathrm{log}(8)-\mathrm{log}(5))}{\mathrm{log}(8)+\mathrm{log}(5)}$$

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.

$${8}^{x-2}\times {5}^{x+2}=1$$

This one according to wolfram alpha it has nice solution

$$x=\frac{2(\mathrm{log}(8)-\mathrm{log}(5))}{\mathrm{log}(8)+\mathrm{log}(5)}$$

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.

giosgi5

Beginner2022-10-22Added 15 answers

$${8}^{x-2}\times {5}^{x+2}={8}^{-2}\cdot {5}^{2}\cdot {8}^{x}\cdot {5}^{x}=\frac{{5}^{2}}{{8}^{2}}\cdot {40}^{x}.$$

That is equal to $1$ precisely if

$${40}^{x}=\frac{64}{25}$$

and that holds only if

$$x={\mathrm{log}}_{40}\frac{64}{25}.$$

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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