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OpenStudy (anonymous):
LOGARITHMS. 12^(x-2)=20 Round To The Nearest Ten-Thousandths
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OpenStudy (aum):
\(\large 12^{x-2} = 20\) Take logarithm on both sides: \(\large \log(12^{x-2}) = \log(20)\) \(\large (x-2)\log(12) = \log(20)\) Can you solve for x?
OpenStudy (anonymous):
Doesnt It Become Like 2xlog12=log20?
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