How do I simplify $\frac{\log_7 32}{\log_7 8\cdot\sqrt2}$?

Question

So far I have got $\log_7 2^5 - \log_7 2^3 + \log_7 2^{1/2}$ and am unable to proceed. Am I, on the right track so far? and how do I proceed? thank you!

Answer 1

Your expansion is not correct. If $b, x, y > 0$, with $b \neq 1$, then \begin{align*} \log_b (xy) & = \log_b x + \log_b y\\ \log_b \left(\frac{x}{y}\right) & = \log_b x - \log_b y \end{align*} However, you have $$\frac{\log_7 32}{\log_7 8\sqrt{2}}$$ which is a quotient of logarithms, not the logarithm of a quotient, so you cannot apply those rules here.

However, your observations that $32 = 2^5$, $8 = 2^{3}$, and $\sqrt{2} = 2^{1/2}$ would be useful if we were working with logarithms to the base $2$. To do so, we use the Change of Base Formula. Let $a, b, x > 0$, with $a, b \neq 1$. Then $$\log_a x = \frac{\log_b x}{\log_b a}$$ By setting $a = 7$ and $b = 2$, we obtain \begin{align*} \frac{\log_7 32}{\log_7 8\sqrt{2}} & = \frac{\dfrac{\log_2 32}{\log_2 7}}{\dfrac{\log_2 8\sqrt{2}}{\log_2 7}}\\ & = \frac{\log_2 32}{\log_2 8\sqrt{2}}\\ & = \frac{\log_2 2^5}{\log_2 2^32^{1/2}}\\ & = \frac{5}{\log_2 2^{7/2}}\\ & = \frac{5}{\frac{7}{2}}\\ & = \frac{10}{7} \end{align*}

Answer 2

You can use the formula

$$\frac{\log_{a}b}{\log_{a}c}=\log_{c}b$$ $$\frac{\log_{7}32}{\log_{7}8}\cdot \sqrt{2}=\left(\log_{8}32\right)\cdot \sqrt{2}=\frac{5\sqrt{2}}{3}$$

Or If you are looking for this

$$\frac{\log_{7}32}{\log_{7}8\sqrt{2}} =\left(\log_{8\sqrt{2}}32\right)=\frac{5}{\frac{7}{2}}=\frac{10}{7}$$