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Prove that....
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\[\huge \log 540 = 2\log2 + 3\log3 + \log 5 \]
I got what you are saying
\(\Large a\log b = \log b^a\)
finally! :P
so 2 log 2 = ... 3 log 3 = ...
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log ( 2 * 27 * 5)
shouldn't there be 4 instead of 2 ?
yes sorry typying mistake
log ( 4 * 27 * 5)
then 4*27*5 is indeed 540
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yes absolutely after that!
any doubts ?
the question is not completed
i mean the answer isn't i think so we have to prove
2log2+3log3+log5
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we proved it...we took the right side, and proved it to left side...
SO, log 2^2 + log 3^3 + log5 2log2 + 3log3 + log5
yes thank you !!
welcome ^_^
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