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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:
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.