$a^{\log_b(c)} = c^{\log_b(a)}$

Question

I'm not sure how to start. My questions is how do you prove:

$$a^{\log_b(c)} = c^{\log_b(a)}$$ where $a,b,c > 0$.

Hint: take log on both sides – Secret Math Jun 15 '13 at 20:17 — Secret Math, Jun 15 '13 at 20:17

score 0 · Answer 1 · answered Jun 15 '13 at 20:15

0

Hint Use that $$\log_b c=\frac{\log c}{\log b}$$

Here $\log$ is the natural logarithm.

answered Jun 15 '13 at 20:15

Pedro

score 0 · Answer 2 · answered Jun 15 '13 at 20:15

0

As $e^{\log_ea}=a$ and $\log_bm=\frac{\log m}{\log b}$

$$a^{\log_bc}=(e^{\log_ea})^{\log_bc}=e^{\log_ea \cdot \log_bc}=e^{\log_ec\cdot \log_ba}=(e^{\log_ec})^{\log_ba}=c^{\log_ba}$$

answered Jun 15 '13 at 20:15

lab bhattacharjee

How do you calculate your third step to your fourth? – user82627 Jun 15 '13 at 21:20
@user82627, As $$\log_bm=\frac{\log m}{\log b}, \log _ea\cdot \log_bc=\frac{\log a}{\log e}\cdot \frac{\log c}{\log b}=\frac{\log c}{\log e}\cdot \frac{\log a}{\log b}=\log_ec\cdot\log_ba $$ – lab bhattacharjee Jun 16 '13 at 05:05

score 0 · Answer 3 · answered Jun 15 '13 at 22:52

0

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

answered Jun 15 '13 at 22:52

Meow

3 Answers3