Raul Walker

Answered

2022-07-14

How do you solve a logarithm with a non-integer base?

How would one calculate the log of a number where the base isn't an integer (in particular, an irrational number)? For example:

${0.5}^{x}=8\text{(where}x=-3\text{)}$

${\mathrm{log}}_{0.5}8=-3$

How would you solve this, and how would this work for an irrational base (like $\sqrt{2}$)?

How would one calculate the log of a number where the base isn't an integer (in particular, an irrational number)? For example:

${0.5}^{x}=8\text{(where}x=-3\text{)}$

${\mathrm{log}}_{0.5}8=-3$

How would you solve this, and how would this work for an irrational base (like $\sqrt{2}$)?

Answer & Explanation

vrtuljakc6

Expert

2022-07-15Added 16 answers

Let's rewrite this in a different way:

${0.5}^{x}=8$

Take the logarithm with respect to any base $a>0$

${\mathrm{log}}_{a}({0.5}^{x})={\mathrm{log}}_{a}8$

which becomes

$x{\mathrm{log}}_{a}0.5={\mathrm{log}}_{a}8$

or

$x=\frac{{\mathrm{log}}_{a}8}{{\mathrm{log}}_{a}0.5}$

You would stop here weren't from the fact that $8={2}^{3}$ and $0.5={2}^{-1}$ so.

$x=\frac{{\mathrm{log}}_{a}8}{{\mathrm{log}}_{a}0.5}=\frac{{\mathrm{log}}_{a}{2}^{3}}{{\mathrm{log}}_{a}{2}^{-1}}=\frac{3{\mathrm{log}}_{a}2}{-{\mathrm{log}}_{a}2}=-3$

You need to compute no logarithm, actually.

${0.5}^{x}=8$

Take the logarithm with respect to any base $a>0$

${\mathrm{log}}_{a}({0.5}^{x})={\mathrm{log}}_{a}8$

which becomes

$x{\mathrm{log}}_{a}0.5={\mathrm{log}}_{a}8$

or

$x=\frac{{\mathrm{log}}_{a}8}{{\mathrm{log}}_{a}0.5}$

You would stop here weren't from the fact that $8={2}^{3}$ and $0.5={2}^{-1}$ so.

$x=\frac{{\mathrm{log}}_{a}8}{{\mathrm{log}}_{a}0.5}=\frac{{\mathrm{log}}_{a}{2}^{3}}{{\mathrm{log}}_{a}{2}^{-1}}=\frac{3{\mathrm{log}}_{a}2}{-{\mathrm{log}}_{a}2}=-3$

You need to compute no logarithm, actually.

invioor

Expert

2022-07-16Added 3 answers

This is the base changing formula :

${\mathrm{log}}_{a}(x)=\frac{{\mathrm{log}}_{b}(x)}{{\mathrm{log}}_{b}(a)}$

${\mathrm{log}}_{a}(x)=\frac{{\mathrm{log}}_{b}(x)}{{\mathrm{log}}_{b}(a)}$

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