What is the solution for 12^x = 18?

[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.


Can someone help please?

Cheers
Vivek

 

[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.

 

[maverick280857]

Thanks Shmoe.