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Write as a single log: (log3^x + log3^12) - log3^2
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Hint: logA+logB-logC = log((A*B)/C)
use laws of logs: log a + log b = log ab log a - log b = log a/b log^(na) = n log a
True. Just a bit of adjustment, right @cwrw238? log a^n = n log a
yea - sorry log a^n = n log a
so it is log(3x*3x^12)(3^2)
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Hmm, not quite. log3^x + log3^12- log3^2 Combining the first two, log(3^x*3^12)-log3^2 Then combining those gives: log((3^x*3^12)/3^2)
|dw:1344025768507:dw| =(x+12)/2
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