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Difference between revisions of "Log Sums Proof"

(Created page with "The sum for logs with the same base goes as the following: <math>\log_{n}a+\log_{n}b=\log_{n}{ab}</math>. But how is this property true? Make <math>\log_{n}a=x</math> and <...")
 
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[[Category:Proofs]]

Revision as of 17:49, 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 $m=\log_{n}c$).


Made by AM24

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