Changing the base of the logarithm

Using the calculator we can determine the logarithms with bases $10$ and $e$ , e.g. $\log(5)$ and $\ln(5)$ , but what about other bases? For example, what is

\log_2(5)=?

Actually, it is possible to convert the base $2$ to a base $10$ or a base $e$ , and then we can use the calculator. Indeed, we have the following useful theorem:

Theorem 1: Change of base (logarithm)

Proof

Here is the proof. We just show that

\log_b(c)=\frac{\log(c)}{\log(b)}

The proof for the $\ln$ is similar. To make things a bit less abstract, let us insert a number of $c$ , e.g. $c=7$ . So can we show that

\log_b(7)=\frac{\log(7)}{\log(b)}

This is correct if

\log(b)\cdot\log_b(7)=\log(7)

and this is correct if

10^{\log(b)\cdot \log_b(7)} = \underbrace{10^{\log(7)}}_{=7}

The left side can be written as

\left(10^{\log(b)}\right)^{\log_b(7)}=b^{\log_b(7)}=7

So we see that the left side equals the right side.

And this concludes the proof!

It is

\log_b(c)=\frac{\log(c)}{\log(b)}=\frac{\ln(c)}{\ln(b)}

Thus, applying this conversion, we have

\log_2(5)=\frac{\log(5)}{\log(2)}=\frac{0.698...}{0.301...}=2.321...

or if you prefer the natural logarithm

\log_2(5)=\frac{\ln(5)}{\ln(2)}=\frac{1.609...}{0.693...}=2.321...

Exercise 1

Without calculator, find a lower and upper bound (both integers) for the logarithm below. Verify your estimate using the calculator.

$\log_8(20)$
$\log_5(124.5)$
$\log_2(0.121)$

Solution

Estimate: We have to find a number $a$ such that $8^a\approx 20$ . $a=1$ is too low, $a=2$ is too large, so $\log_8(20)\in \underline{]1,2[}$ . The exact value is
$\log_8(20)=\frac{\log(20)}{\log(8)}=\underline{1.440...}$
Estimate: We have to find a number $a$ such that $5^a\approx 124.5$ . $a=3$ is too high, $a=2$ is too low, so $\log_5(124.5)\in \underline{]2,3[}$ . The exact value is
$\log_5(124.5)=\frac{\log(124.5)}{\log(5)}=\underline{2.997...}$
Estimate: We have to find a number $a$ such that $2^a\approx 0.121$ . For $a=-3$ it is $2^{-3}=\frac{1}{8}=0.125$ is too hight, so $a=-3$ is the upper bound, and $a=-$ 4 therefore the lower bound. Thus $\log_2(0.121)\in \underline{]-4,-3[}$ . The exact value is
$\log_2(0.121)=\frac{\log(0.121)}{\log(2)}=\underline{-3.046...}$