How do you find the exact value of a logarithm with a radical in the base?

Question

I'm struggling to find a method for evaluating $\log_{5\sqrt2} 50$ (or ${\log50}\over{\log5\sqrt2}$) without using a calculator. When using a calculator, I am given an exact value of 2, but I can't figure out why without one? Does it have to do with the $\sqrt2$ being the same as $2^\frac12$? Is there some kind of rule/property that I'm forgetting about? Even when rewriting in exponential form (${(5\sqrt2)}^x = 50$), I can't come up with anything. I'm a college student in precalculus and can't find any explanation in the textbook.

Answer 1

I look at $5\sqrt2$ and $50$ in the same formula and it is almost an automatic reflex to observe that $50$ is the square of $5\sqrt2$. This comes from habitually looking for such relations between numbers.

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But since you have not had a chance to develop such a habit, here's a more methodological approach, based on prime factors.

$$ 50 = 2 \cdot 5^2. $$

$$ 5\sqrt2 = 2^{1/2} \cdot 5. $$

Therefore

$$ \log(50) = \log(2) + 2\log(5). $$ $$ \log(5\sqrt2) = \frac12\log(2) + \log(5). $$

Can you find the ratio $\dfrac{\log(50)}{\log(5\sqrt2)}$ from that information?

Answer 2

$10^x = 5\sqrt 2 \implies 10^{2x} = 50$

$\log_{5 \sqrt 2} 50 = \frac{\log 50}{\log 5\sqrt 2 } = \frac{2x}{x} = 2$

Answer 3

Let $$\log_{5\sqrt{2}} 50 = x$$
Then, by the definition of the logarithm, $$(5\sqrt{2})^x = 50$$ If we factor $50$ into primes, we obtain $50 = 5^2 \cdot 2$. Since $2 = \sqrt{2}^2$, $50 = 5^2 \cdot 2 = 5^2 \cdot \sqrt{2}^2 = (5\sqrt{2})^2$. Thus, \begin{align*} (5\sqrt{2})^x & = (5\sqrt{2})^2\\ \log_{5\sqrt{2}} (5\sqrt{2})^x & = \log_{5\sqrt{2}} (5\sqrt{2})^2\\ x & = 2 \end{align*}