# Why is $a^{\log_b n}$ the same as $n^{\log_b a}$?

I was watching video Lec 2 | MIT 6.046J / 18.410J Introduction to Algorithms (SMA 5503), Fall 2005, where professor Erik Demaine said that $$a^{\log_b n}$$ is the same as $$n^{\log_b a}$$. Can someone please explain me why bringing "the $$n$$ downstairs" and "the $$a$$ upstairs" is possible?

• It's a property of logarithm. I hope this link will help. May 5 at 16:20
• Thanks @Russel. May 6 at 8:10

Take $$\log_b$$, and you get $$\log_b n \cdot \log_b a$$ on both sides.

We can say some number $$a = b^{\log_b(a)}$$, because $$\log_b(a)$$ tells us to how much power we need to raise $$x$$ to get $$a$$ and then we actually raise it to get $$a$$.

Then, $$a^k = (b^{\log_b(a)})^k = b^{k\log_b(a)}$$

If $$k = \log_b(n)$$, then $$a^{\log_b(n)} = b^{\log_b(n)\log_b(a)}$$

Notice that $$n = b^{\log_b(n)}$$, then:

$$a^{\log_b(n)} = (b^{\log_b(n)})^{\log_b(a)} = n^{\log_b(a)}$$ $$a^{\log_b(n)} = n^{\log_b(a)}$$

$$p^{\log(q)}=e^{\log(p)\log(q)}=q^{\log(p)}$$ holds (also for other bases).

The first result that will be useful is $$\ln x^r = r \ln x$$, where $$\ln$$ is the natural logarithm. When the base of the logarithm changes to a different number $$a$$, the logarithm can be rewritten as a ratio natural logarithms: $$\log_a x = \frac{\ln x}{\ln a}$$. This can be proved by starting with $$y=\log_a x$$, or its equivalent equation $$x=a^y$$ and taking the natural logarithm of both sides. We get $$\ln x = \ln a^y$$, and applying the first result above gives $$\ln x=y \ln a$$, from which we get $$y = \frac{\ln x}{\ln a}$$, i.e. that $$\log_a x = \frac{\ln x}{\ln a}$$.

To show that $$a^{\log_b n}$$ and $$n^{\log_b a}$$ are equal, we show their natural logarithms are equal. After taking logarithms, we can apply the first result above. We get that the logarithm of the first expression is $$\log_b n \ln a = \frac{\ln n }{\ln b} \ln a$$. The logarithm of the second expression is $$\log_b a \ln n = \frac{\ln a }{\ln b} \ln n$$. These two ratios are both equal to $$\frac{\ln a \ln n}{\ln b}$$.