# 12^x = 18

1. Jun 2, 2005

### maverick280857

Hi everyone. Here's a simple problem I need help with:

Find x such that $12^x = 18$

From one point of view, $x = log(18)/log(12)$ and the problem is solved.

However, if we write 12 as $3*2^2$ and 18 as $3^2*2$ then,

$(2 * 3^2) = 3^x * 2^{2x}$

and hence by the uniqueness of prime factorization (in particular that of the exponents of the prime factors),

x = 2
and 2x = 1

but these equations do not have a consistent solution. I think the error is in the second reasoning.

Cheers
Vivek

Last edited: Jun 2, 2005
2. Jun 3, 2005

### shmoe

You can only guarantee unique factorization if you have integral exponents. When you try to invoke the fundamental theorem of arithmetic you would also need the assumption that the exponents x and 2x are integers. Your contradiction only tells you that there is no integer x that satisfies the original equation.

3. Jun 3, 2005

### maverick280857

Thanks Shmoe.