# Thread: Exponents solving for n!

1. Good post? |

## Exponents solving for n!

I struggle with the following equation that hast to be solved for n:

9^7+9^7+9^7 = 27^n+1

I tried to factor the numbers, but nevertheless I didn't get the right result.
3(9^7)= (3*9)^n+1
3(3^7*3^7)= (3*3*3)^n+1
3^15=3^3n+3

15=3n+3
n=4

But this solution is wrong! Why?

2. Good post? |
3^15=3^3n+3^0
15=3n+0
15=3n
n=5

3. Good post? |
I am not sure whether the equation is:

1) 9^7+9^7+9^7 = 27^(n+1)

or

2) 9^7+9^7+9^7 = (27^n)+1

If it is case (1), then n= 4 is correct...No doubt abt it..!!

In case (2), it is not simple equation where u can add indices and equate them getting the answer. The proper way to solve that would be using logarithms which is beyong the scope of GMAT.
Yeah, we are sure that the exact value of 'n' here wont be a whole number.
WIth approximation, we can say that it will around 4.8 or 4.9 ~ 5.
(Since the value of 3^15 will be very large and will almost unchanged even if we subtract 1 from it. Thus, we can say that 3^15 will be almost equal to 3^3n. Therefore n = 5).

GMAT02,

Indices dont have associate property. I mean;

A^p = A^q + A^r cant be made to p = q + r

So u cant do:
3^15=3^3n+3^0
15=3n+0

Thanks and regards,
Vivek

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