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Log Sums Proof: Difference between revisions

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Made by [https://artofproblemsolving.com/wiki/index.php/User:Am24|AM24]
Made by [https://artofproblemsolving.com/wiki/index.php/User:Am24 AM24]


[[Category:Proofs]]
[[Category:Proofs]]

Latest revision as of 21:22, 26 August 2025

The sum for logs with the same base goes as the following:

$\log_{n}a+\log_{n}b=\log_{n}{ab}$.

But how is this property true?

Make $\log_{n}a=x$ and $\log_{n}b=y$. Now we can see (because of logs work) $n^x=a \text{and}n^y=b$. If we were to multiply these two values together we would get $n^{x+y}=ab$. $x+y$ is the sum of our two logs, so were finished. WRONG. We still have a crucial step.

Just for neatness we will sub in $x+y$ with $S$ because that's our sum.

$n^S=ab \implies S=\boxed{\log_{n}ab}$ (whenever you have $n^m=c$ then $m=\log_{n}c$).


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