Simplifying logarithms and changing base

Question

I have been asked to find the value of

$8^{\log_{2} 5}$

I understand that I could proceed to turn this into

$\log_{8} x = \log_{2} 5$

Where do I go from there? I assumed changing both to the same base, but I'm not sure how to do so or what to do after having the same base for both sides.

Thank you!

Answer 1

Since $8=2^3$ and $2^{log_2 5}=5$, we have $$8^{log_2 5}=(2^3)^{log_2 5}=(2^{log_2 5})^3=5^3=125$$

Answer 2

Hint $8 = 2^3$ and whenever $x = y > 0$ you can deduce $\log_2(x) = \log_2(y)$.

Answer 3

Note that $\log_b a = \dfrac{\log_c a}{\log_c b}$ so in this case $$\log_2 5 = \frac{\log_8 5}{\log_8 2} = \frac{\log_8 5}{\log_8 8^{1/3}} = 3\log_8 5 = \log_8 5^3$$

So that $$8^{\displaystyle \log_2 5} = 8^{\displaystyle \log_8 5^3} = 5^3 = 125$$

Note that $a^{\displaystyle \log_a x} = x$ by virtue of the composition of $a^x$ with its inverse $\log_a x$, giving $\mathrm{id}$.

Answer 4

Let $\log_25=y\implies2^y=5$

Now $8^{\log_25}=(2^3)^y=(2^y)^3=5^3$

See Laws of Logarithms and Exponent Combination Laws

Answer 5

$\log_8x=\log_25 \implies \frac{1}{3} \log_2 x = \log_2 5 \implies \log_2 x = 3\log_2 5 \implies \log_2 x = \log_2 125$

Since $\log_k x$ is a bijective function over $\mathbb R^+$, $k \in (0,1)\cup(1,+\infty)$

$x = 125$