Meldeaktezl

2022-09-09

Quick Log problem?

${2}^{({x}^{3})}={3}^{({x}^{2})}$

Solve for x

I'm pretty sure I use logs to solve this, but how? to what base? I'm kinda lost.. Thanks

${2}^{({x}^{3})}={3}^{({x}^{2})}$

Solve for x

I'm pretty sure I use logs to solve this, but how? to what base? I'm kinda lost.. Thanks

Haylee Branch

Beginner2022-09-10Added 7 answers

Taking logs to any base,

$${x}^{3}\mathrm{log}2={x}^{2}\mathrm{log}3\text{},$$

and therefore either $x=0$ or $x=(\mathrm{log}3)/(\mathrm{log}2)$. Note that by the "change of base" formula, the last expression is the same no matter what base you are using.

$${x}^{3}\mathrm{log}2={x}^{2}\mathrm{log}3\text{},$$

and therefore either $x=0$ or $x=(\mathrm{log}3)/(\mathrm{log}2)$. Note that by the "change of base" formula, the last expression is the same no matter what base you are using.

ivybeibeidn

Beginner2022-09-11Added 2 answers

Take natural log

$${2}^{{x}^{3}}={3}^{{x}^{2}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{3}\mathrm{ln}2={x}^{2}\mathrm{ln}3\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{2}(x\mathrm{ln}2-\mathrm{ln}3)=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{1,2}=0,\text{}{x}_{3}=\frac{\mathrm{ln}3}{\mathrm{ln}2}={\mathrm{log}}_{2}3$$

$${2}^{{x}^{3}}={3}^{{x}^{2}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{3}\mathrm{ln}2={x}^{2}\mathrm{ln}3\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{2}(x\mathrm{ln}2-\mathrm{ln}3)=0\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{1,2}=0,\text{}{x}_{3}=\frac{\mathrm{ln}3}{\mathrm{ln}2}={\mathrm{log}}_{2}3$$

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