Logarithm Base Changing Formula Proof

In a logarithmic expression, it is possible to change base using algebraic manipulation.  For example, we can change

$log_416$ to $\dfrac{\log_216}{\log_24}$.

In this post, we are going to prove why it is possible to do such algebraic manipulation. The change of base above can be generalized as

$\log_ab = \dfrac{\log_cb}{\log_ca}$.

Theorem

$\log_ab = \dfrac{log_cb}{\log_ca}$.

Proof

If we let $\log_ab = x$, then by definition, $a^x = b$.

Now, take the logarithm to the base $c$ of both sides. That is

$log_c a^x = \log_cb$.

Simplifying the exponent, we have

$x \log_ca = \log_cb$.

Now, since $a \neq 1$, $\log_ca \neq 0$.

Therefore,

$x = \dfrac{\log_cb}{\log_ca}$

Thus,

$\log_ab = \dfrac{log_cb}{\log_ca}$

Year 2015 in Review – Complete List of Posts

It’s the end of the year again, so let’s look at what we have learned so far in 2015. Below is the complete list of posts of mathematical proofs this year. Enjoy learning!

Year 2015 in Review – Complete List of Posts

You may also want to visit the Post List Page to explore the complete list of posts of Proofs from the Book.

Proof that the nth root of a times nth root of b is equal to nth root of ab

Consider the following multiplications.

$\sqrt{4} \times \sqrt{16} = 2 \times 4 = 8$
$\sqrt{4 \times 16} = \sqrt{64} = 8$.

Also

$\sqrt[3]{27} \times \sqrt[3]{64} = 3 \times 4 = 12$
$\sqrt[3]{27} \times \sqrt[3]{64} = \sqrt[3]{1728} = 12$.

Try also several calculations like these and observe what happens.

From these calculations, we can observe that

$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$

and

$\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$

or, in general

$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$.

We now, prove our conjecture.

Theorem: $\sqrt[n]{a} \sqrt[n]{b} = \sqrt[n]{ab}$

Proof

If we multiply the expression on the left hand side by itself n times, we have

$(\sqrt[n]{a} \sqrt[n]{b})^n = (\sqrt[n]{a})^n (\sqrt[n]{b})^n = ab$.

On the other hand, a positive number ab has only a single positive root, $\sqrt[n]{ab}$.

Therefore,

$\sqrt[n]{a} \sqrt[n]{b} = \sqrt[n]{ab}$.

This completes our proof.