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

Question

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

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

$$\log_{0.5}8 = -3$$

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

Answer 1

Let's rewrite this in a different way: $$ 0.5^x=8 $$ Take the logarithm with respect to any base $a$ ($a>0$, $a\ne1$): $$ \log_a(0.5^x)=\log_a8 $$ which becomes $$ x\log_a 0.5=\log_a 8 $$ or $$ x=\frac{\log_a 8}{\log_a 0.5} $$ You would stop here weren't from the fact that $8=2^3$ and $0.5=2^{-1}$, so $$ x=\frac{\log_a 8}{\log_a 0.5}=\frac{\log_a 2^3}{\log_a 2^{-1}}= \frac{3\log_a 2}{-\log_a 2}=-3 $$ You need to compute no logarithm, actually.

Answer 2

rewrite your equation in the form $2^{-x}=2^{3}$

Answer 3

This is the base changing formula : $$\log_{a}(x) = \frac{\log_{b}(x)}{\log_{b}(a)}$$

Answer 4

Yet another attempt to help. Let’s look at a somewhat different case. How do you calculate $\log_{3.8}(8)$? Again as egreg says, rewrite as $3.8^x=8$, take the (ordinary) log of both sides to get $x\log_{10}(3.8)=\log_{10}(8)$, and $x=\log_{10}(8)/\log_{10}(3.8)$

If you want $\log_{\sqrt2}(8)$, it’s the same thing, only you can shortcut matters early: $8={\sqrt2\,}^x=(2^{1/2})^x=2^{x/2}$; knowing that $8=2^3$, you see that $x=6$, $\log_{\sqrt2}(8)=6$.

Answer 5

You can use natural logarithms, because for all $a$ and $b$ : $\log_a(b)=\frac{\ln(b)}{\ln(a)}$.