Help with specific question on logs

[Flyboy10]

If [(3^n)-1]/2 = 900

show that n = [log(900) / log(3)] +1

This is from a past A/S Level C2 paper and I can't get it to come out the way they want.

The closest I can get is that n= [log((900 x 2) + 1)] / log(3)


Can anyone show me how to put in in the way they asked for?

:duck:

[jemidiah]

The given answer is incorrect (though the expression you've given is correct).

Suppose to the contrary n = [log(900) / log(3)] + 1. Then...
3^n = 3^(log(900)/log(3) + 1) = 3*3^(log_3 (900)) = 3*900 = 2700
=> [(3^n)-1]/2 = [2699]/2 = 1349.5 != 900

I verified this on a calculator as well. Perhaps there's a typo, you misread a grouping symbol, or the answer is just plain wrong.