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I'm being required to prove the power property of logarithms, but in the same way that my teacher did in our precalc class two weeks ago (which I do not remember), which was by simplifying only a single side of the equation to be equal to the other.

I've been trying to wrap my head around this and have found that it is relatively easy to prove the theorem by either setting the logarithm with the power to equal a variable and changing it to and later from exponential form, and I also managed to make the two sides equal to each other by taking a (my example base) to the power of each side, then simplifying.

I cannot however figure out how I am suppose to proceed when only working on one side of the equation. The hint suggests to use the change of base formula, and I have a suspicion that this involves multiplying exponents together, but I cannot figure out how.

Can someone help me out with this?

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  • $\begingroup$ My apologies, I meant the power property, which is that the coefficient in front of a logarithm can also be placed as the exponent to the value inside of the logarithm. I have fixed where I wrote it incorrectly. $\endgroup$ – NuffsaidM8 Nov 14 '17 at 23:15
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You want to show that $$\log_a(b^c)=c\log_ab$$

Change the base on the lefthand side to base $b$, which gives $$\log_a(b^c)=\frac{\log_b(b^c)}{\log_ba}=\frac{c}{\color{red}{\log_ba}}=\frac{c}{\color{blue}{\log_aa/\log_ab}}=c\log_ab$$

where I used the change of base formula a second time to rewrite the red expression (in base $b$) as the blue expression (in base $a$).

(The proof uses the facts that $\log_b(b^c)=c$ and $\log_aa=1$ by definition.)

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  • $\begingroup$ That is exactly what I wanted. I started with the change of base formula, but was unable to proceed because I didn't divide by the right variable to cancel things and continue. Many thanks. $\endgroup$ – NuffsaidM8 Nov 14 '17 at 23:21
  • $\begingroup$ @NuffsaidM8: You're welcome. You can show your thanks on this site by upvoting and by clicking the check mark to mark the question as resolved. $\endgroup$ – symplectomorphic Nov 14 '17 at 23:22

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