Kubaczynjo

2022-08-30

Logarithm problems with different bases

1.${\mathrm{log}}_{a}b\times {\mathrm{log}}_{b}a=$?

2.${\mathrm{log}}_{a}b+{\mathrm{log}}_{b}a=\sqrt{29}$

What is ${\mathrm{log}}_{a}b-{\mathrm{log}}_{b}a=$?

3.

What is b in the following:

${\mathrm{log}}_{b}3+{\mathrm{log}}_{b}11+{\mathrm{log}}_{b}61=1$

and

4.

$\frac{1}{lo{g}_{2}x}+\frac{1}{lo{g}_{25}x}-\frac{3}{{\mathrm{log}}_{8}x}=\frac{1}{{\mathrm{log}}_{b}x}$

What is b?

Can anyone help me solve these?

1.${\mathrm{log}}_{a}b\times {\mathrm{log}}_{b}a=$?

2.${\mathrm{log}}_{a}b+{\mathrm{log}}_{b}a=\sqrt{29}$

What is ${\mathrm{log}}_{a}b-{\mathrm{log}}_{b}a=$?

3.

What is b in the following:

${\mathrm{log}}_{b}3+{\mathrm{log}}_{b}11+{\mathrm{log}}_{b}61=1$

and

4.

$\frac{1}{lo{g}_{2}x}+\frac{1}{lo{g}_{25}x}-\frac{3}{{\mathrm{log}}_{8}x}=\frac{1}{{\mathrm{log}}_{b}x}$

What is b?

Can anyone help me solve these?

betoosolis7i

Beginner2022-08-31Added 12 answers

The way to start all of these and turn them into simple algebra is that ${\mathrm{log}}_{a}b=\frac{{\mathrm{log}}_{x}b}{{\mathrm{log}}_{x}a}$ Using that formula, all of these become basic algebra. Give it a try and comment what you get.

Nezveda6q

Beginner2022-09-01Added 7 answers

2 . let $x={\mathrm{log}}_{a}b=\frac{1}{{\mathrm{log}}_{b}a}$

${\mathrm{log}}_{a}b+\frac{1}{{\mathrm{log}}_{a}b}=\sqrt{29}$

Then ${\mathrm{log}}_{a}b$ and ${\mathrm{log}}_{a}b$ are solutions of the following equation

$x+\frac{1}{x}=\sqrt{29}$

${x}_{1}=\frac{\sqrt{29}-5}{2}={\mathrm{log}}_{b}a$

${x}_{2}=\frac{\sqrt{29}+5}{2}={\mathrm{log}}_{a}b$

(We assume that $a<b$ ) Thus ${\mathrm{log}}_{a}b-{\mathrm{log}}_{b}a=\frac{\sqrt{29}+5}{2}-\frac{\sqrt{29}-5}{2}=5$

4.

$\frac{1}{lo{g}_{2}x}+\frac{1}{lo{g}_{25}x}-\frac{3}{{\mathrm{log}}_{8}x}=\frac{1}{{\mathrm{log}}_{b}x}$

Using formula ${\mathrm{log}}_{a}b=\frac{1}{{\mathrm{log}}_{b}a}$ we have

${\mathrm{log}}_{x}2+{\mathrm{log}}_{x}25-3lo{g}_{x}8={\mathrm{log}}_{x}b$

Use formulas

${\mathrm{log}}_{x}a+{\mathrm{log}}_{x}c={\mathrm{log}}_{x}ac$

and

$n{\mathrm{log}}_{x}a={\mathrm{log}}_{x}{a}^{n}$

${\mathrm{log}}_{a}b+\frac{1}{{\mathrm{log}}_{a}b}=\sqrt{29}$

Then ${\mathrm{log}}_{a}b$ and ${\mathrm{log}}_{a}b$ are solutions of the following equation

$x+\frac{1}{x}=\sqrt{29}$

${x}_{1}=\frac{\sqrt{29}-5}{2}={\mathrm{log}}_{b}a$

${x}_{2}=\frac{\sqrt{29}+5}{2}={\mathrm{log}}_{a}b$

(We assume that $a<b$ ) Thus ${\mathrm{log}}_{a}b-{\mathrm{log}}_{b}a=\frac{\sqrt{29}+5}{2}-\frac{\sqrt{29}-5}{2}=5$

4.

$\frac{1}{lo{g}_{2}x}+\frac{1}{lo{g}_{25}x}-\frac{3}{{\mathrm{log}}_{8}x}=\frac{1}{{\mathrm{log}}_{b}x}$

Using formula ${\mathrm{log}}_{a}b=\frac{1}{{\mathrm{log}}_{b}a}$ we have

${\mathrm{log}}_{x}2+{\mathrm{log}}_{x}25-3lo{g}_{x}8={\mathrm{log}}_{x}b$

Use formulas

${\mathrm{log}}_{x}a+{\mathrm{log}}_{x}c={\mathrm{log}}_{x}ac$

and

$n{\mathrm{log}}_{x}a={\mathrm{log}}_{x}{a}^{n}$

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