Salvador Bush

Answered

2022-07-12

Something to the power of a logarithm

This is probably a very obvious question, but here goes...

An answer in my textbook claims that

${3}^{\mathrm{log}n}={n}^{\mathrm{log}3}$

and that

$4{n}^{2}(3/4{)}^{\mathrm{log}n}=4{n}^{\mathrm{log}3}$

Why, using more basic laws, is this the case?

(Unfortunately Google confuses this question with changing bases, exponentiation being the inverse of log (which is of course related), and similar matters.)

This is probably a very obvious question, but here goes...

An answer in my textbook claims that

${3}^{\mathrm{log}n}={n}^{\mathrm{log}3}$

and that

$4{n}^{2}(3/4{)}^{\mathrm{log}n}=4{n}^{\mathrm{log}3}$

Why, using more basic laws, is this the case?

(Unfortunately Google confuses this question with changing bases, exponentiation being the inverse of log (which is of course related), and similar matters.)

Answer & Explanation

alomjabpdl0

Expert

2022-07-13Added 12 answers

Recall that $3={b}^{{\mathrm{log}}_{b}3}$

Therefore ${3}^{{\mathrm{log}}_{b}n}={\textstyle (}{b}^{{\mathrm{log}}_{b}3}{{\textstyle )}}^{{\mathrm{log}}_{b}n}={b}^{({\mathrm{log}}_{b}3)({\mathrm{log}}_{b}n)}={\textstyle (}{b}^{{\mathrm{log}}_{b}n}{{\textstyle )}}^{{\mathrm{log}}_{b}3}={n}^{{\mathrm{log}}_{b}3}$

Therefore ${3}^{{\mathrm{log}}_{b}n}={\textstyle (}{b}^{{\mathrm{log}}_{b}3}{{\textstyle )}}^{{\mathrm{log}}_{b}n}={b}^{({\mathrm{log}}_{b}3)({\mathrm{log}}_{b}n)}={\textstyle (}{b}^{{\mathrm{log}}_{b}n}{{\textstyle )}}^{{\mathrm{log}}_{b}3}={n}^{{\mathrm{log}}_{b}3}$

cooloicons62

Expert

2022-07-14Added 4 answers

Hint : $a=b$ implies $\mathrm{log}a=\mathrm{log}b$ and $\mathrm{log}{a}^{b}=b\mathrm{log}a$

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