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2025 AMC 12A Problems/Problem 7: Difference between revisions

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Solution 1:
Solution 1:
Taking the logarithm of both sides in the first equation, we have:
Taking the logarithm of both sides and using n = 5, we have:


<imath>log</imath> <imath> kn^am^b</imath>
<imath>log</imath> <imath> k5^am^b</imath> = $4+2logm

Revision as of 16:36, 6 November 2025

Solution 1: Taking the logarithm of both sides and using n = 5, we have:

$log$ $k5^am^b$ = $4+2logm

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