orkesruim40

2022-08-19

Prove equality ${a}^{{\mathrm{log}}_{b}c}={c}^{{\mathrm{log}}_{b}a}$

I'm try to prove the equality:

${a}^{{\mathrm{log}}_{b}c}={c}^{{\mathrm{log}}_{b}a}$

I'm having trouble finding information regarding this, also I need to figure out why ${n}^{{\mathrm{log}}_{2}3}$ is better than ${3}^{{\mathrm{log}}_{2}n}$ as a closed form formula for $M(n)$? Thank you!

I'm try to prove the equality:

${a}^{{\mathrm{log}}_{b}c}={c}^{{\mathrm{log}}_{b}a}$

I'm having trouble finding information regarding this, also I need to figure out why ${n}^{{\mathrm{log}}_{2}3}$ is better than ${3}^{{\mathrm{log}}_{2}n}$ as a closed form formula for $M(n)$? Thank you!

yassou1v

Beginner2022-08-20Added 14 answers

Taking the logarithm to base $b$ of both sides, the equation is equivalent to

$({\mathrm{log}}_{b}c)({\mathrm{log}}_{b}a)=({\mathrm{log}}_{b}a)({\mathrm{log}}_{b}c)\text{}.$

I would say ${n}^{{\mathrm{log}}_{2}3}$ is better because it makes it clear that it is $n$ to a constant power. This is of course also true for ${3}^{{\mathrm{log}}_{2}n}$, but it's not so obvious.

$({\mathrm{log}}_{b}c)({\mathrm{log}}_{b}a)=({\mathrm{log}}_{b}a)({\mathrm{log}}_{b}c)\text{}.$

I would say ${n}^{{\mathrm{log}}_{2}3}$ is better because it makes it clear that it is $n$ to a constant power. This is of course also true for ${3}^{{\mathrm{log}}_{2}n}$, but it's not so obvious.

sittesf

Beginner2022-08-21Added 1 answers

For the first part, you could use the fact that

$\begin{array}{rl}{\mathrm{log}}_{b}c& =\frac{{\mathrm{log}}_{a}c}{{\mathrm{log}}_{a}b}\\ & =\frac{{\mathrm{log}}_{a}c}{\frac{{\mathrm{log}}_{b}b}{{\mathrm{log}}_{b}a}}\\ & =({\mathrm{log}}_{a}c)\cdot ({\mathrm{log}}_{b}a)\end{array}$

And then apply it to LHS:

$\begin{array}{rl}{a}^{{\mathrm{log}}_{b}c}& ={a}^{({\mathrm{log}}_{a}c)\cdot ({\mathrm{log}}_{b}a)}\\ & ={\left({a}^{{\mathrm{log}}_{a}c}\right)}^{{\mathrm{log}}_{b}a}\\ & ={c}^{{\mathrm{log}}_{b}a}\end{array}$

$\begin{array}{rl}{\mathrm{log}}_{b}c& =\frac{{\mathrm{log}}_{a}c}{{\mathrm{log}}_{a}b}\\ & =\frac{{\mathrm{log}}_{a}c}{\frac{{\mathrm{log}}_{b}b}{{\mathrm{log}}_{b}a}}\\ & =({\mathrm{log}}_{a}c)\cdot ({\mathrm{log}}_{b}a)\end{array}$

And then apply it to LHS:

$\begin{array}{rl}{a}^{{\mathrm{log}}_{b}c}& ={a}^{({\mathrm{log}}_{a}c)\cdot ({\mathrm{log}}_{b}a)}\\ & ={\left({a}^{{\mathrm{log}}_{a}c}\right)}^{{\mathrm{log}}_{b}a}\\ & ={c}^{{\mathrm{log}}_{b}a}\end{array}$

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