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?

Last edited:


Science Advisor
Homework Helper
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.

Want to reply to this thread?

"12^x = 18" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads