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Solve for x 83=27^x Consider that: log(a^x)=x (times) log(a)
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It looks like you already have your answer there. So just apply it and solve for x.
\[\log(83)=xlog(27)\] \[x=\frac{\log(83)}{\log(27)}\]
on the log(27) part, do you just ignore the x over the 27 ? or where did it goes ?
That's where the rule comes in. if 8=2^3 then log8=3log2
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