tefoldaavv

2022-04-26

How $a}^{{\mathrm{log}}_{b}x}={x}^{{\mathrm{log}}_{b}a$

Leia Wiggins

Beginner2022-04-27Added 18 answers

For any positive $r$ and you have

${r}^{s}={b}^{{\mathrm{log}}_{b}\left({r}^{s}\right)}={b}^{s{\mathrm{log}}_{b}\left(r\right)}.$

So, taking $r=a$ and $s={\mathrm{log}}_{b}\left(x\right)$, we have:

$a}^{{\mathrm{log}}_{b}\left(x\right)}={b}^{{\mathrm{log}}_{b}\left(x\right){\mathrm{log}}_{b}\left(a\right)$

$={b}^{{\mathrm{log}}_{b}\left(a\right){\mathrm{log}}_{b}\left(x\right)}$

$={b}^{{\mathrm{log}}_{b}\left({x}^{{\mathrm{log}}_{b}\left(a\right)}\right)}$

$={x}^{{\mathrm{log}}_{b}\left(a\right)}.$

Gianna Travis

Beginner2022-04-28Added 11 answers

Using log properties, we have

$a}^{{\mathrm{log}}_{b}\left(x\right)}={b}^{{\mathrm{log}}_{b}\left({a}^{{\mathrm{log}}_{b}\left(x\right)}\right\}={b}^{{\mathrm{log}}_{b}\left(x\right){\mathrm{log}}_{b}\left(a\right)}={\left({b}^{{\mathrm{log}}_{b}\left(x\right)}\right)}^{{\mathrm{log}}_{b}\left(a\right)}={x}^{{\mathrm{log}}_{b}\left(a\right)}$

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