log to exponential form, but with number in front of log

Question

So I understand how to put a log equation into exponential form. For example, $y = \log_2(x)$ is $2^y = x.$ However, I don't understand what to do when there is a number in front of $\log$, such as $y = 2\log_2(x)$.

Answer 1

Well what happens when you try? (I will you 3 instead of 2)

Suppose $y = 3 \log_2 x$

So $2^y = 2^{3\log_2 x} = (2^{\log_2 x})^3 = x^3$.

So if $y = k \log_b x$ then $b^y = x^k$.

It's also worth noting that $k \log_b x = \log_b(x^k)$. It's a useful rule to remember and it makes sense when you think about it.

($\log_b x = v \iff b^v = x$

Then $\log_b x^k = \log_b (b^v)^k = \log_b b^{vk} = kv = k \log_b x$)

Knowing that rule makes the process trivial:

$y = 3 \log_2 x = \log_2 x^3$ so $2^y = x^3$.

Answer 2

Lets start with $ y = 2 \log_2(x) $ which is the same as saying $ y = \log_2(x) + \log_2(x) $ . Since both side are equal if we do the same thing to them they will remain equal. So we can both both sides on top of 2.

$ 2 ^ y = 2^{(\log_2(x) + \log_2(x) )} $

Since $ a^b *a^c = a^{(b+c)}$ we can use that to expand it out.

$ 2 ^ y = 2^{\log_2(x)}2^{\log_2(x) )} $

Then using the definition of log to simply.

$ 2 ^ y = xx $

$ 2 ^ y = x^2 $