yet another log prob: solve 2.25 = 2^(n/12) for n

[iamtrojan5]

How would I solve the following for n?

2.25=2^(n/12)

Yes, this is another physics problem...
Thank you for your help!

 

[morson]

Just take the logarithm of base 10 of both sides:

log10(2.25) = log10(2^(n/12))

Use the fact that loga(b^n) = nloga(b) on RHS:

log10(2.25) = n/12 * log10(2)

==> n * log10(2) = 12log10(2.25)

==> n = 12log10(2.25) / log10(2)

Note: you could've taken any base log, but I used 10 for convenience (the log10 function appears on most calculators)