12^x = 18

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


Find x such that [itex]12^x = 18[/itex]

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

However, if we write 12 as [itex]3*2^2[/itex] and 18 as [itex]3^2*2[/itex] then,

[itex](2 * 3^2) = 3^x * 2^{2x}[/itex]

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
 
Last edited:

shmoe

Science Advisor
Homework Helper
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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.
 
Thanks Shmoe.
 

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