Is there a power of 3 and power of 7 that are equal?

Question

I have been trying to find a set of values $(a,b)\in\mathbb Z^+$ which satisfy the following relationship:

$3^a = 7^b$

I haven not been able to find a value that can satisfy this and I also can't find a way disprove it.

Is there a theorem that I am missing?

Answer 1

The theorem you are missing is the fundamental theorem of arithmetic which states that every integer factors uniquely as a product of prime integers.

Answer 2

For fun:

Assume there are $m,n \in \mathbb{Z^+}$, s.t.

$3^n=7^m.$

Euclid's lemma:

If a prime $p$ divides $ab$, $a,b$, integers,

then $p$ divides $a$ or $p$ divides $b$.

$3|3^n$, and $3^n=7^m$, implies $3|7^m.$

But $3$ does not divide $7.$

A contradiction.

Can you show that

$3\not | 7^m$ repeatedly using Euclid's lemma ?

Answer 3

You can prove that differently as well.

Since $a,b$ are integers, $\frac{a}{b}$ is rational by definition.

But:

$$3^a=7^b \to a\log 3 = b\log 7 \to \frac{a}{b} = \frac{\log7}{\log3}$$ Which is an irrational number. Hence, there are not integers $a,b$ such that: $3^a=7^b$