# Logarithm Question With Square Root

There is a problem that I cannot solve:

$$\sqrt{ (\log {8})^2 + \left(\log {\frac {1}{16}}\right)^2}$$

$$A) \sqrt{2} \log {2}$$ $$B) \log {2}$$ $$C) 3\log {2}$$ $$D) 5\log {2}$$ $$E) 5$$

Rules for logaritm:

$$\log_a{b^c} = c.\log_a{b}$$

Therefore inner of the square:

$$\log({2^3})^2 + \log({2^{-4}})^2$$ $$\log({2^6}) + \log({2^{-8}})$$ $$6 \log({2}) -8 \log({2})$$ $$-2 \log({2})$$

Where is my error at the calculation? Thanks in advance.

Mr. Oscar Lenzi has informed me about the solution. Here is the solution: (inner of the square)

$$(\log{2^3})^2 + (\log{2^{-4}})^2$$ $$(3\log{2})^2 + (-4 \log{2})^2$$ $$9 (\log{2})^2 + 16 (\log{2})^2$$ $$25 (\log{2})^2$$

$$\sqrt{25 (\log{2})^2}$$ $$5 \log{2}$$

• Note that $(\log8)^2\ne\log(8^2)$. Commented Mar 7, 2021 at 20:35
You misdid parentheses. Properly, $$(\log 8)^2=(3\log 2)^2=9[(\log 2)^2]$$. Similarly for $$[\log (1/16)]^2$$. With careful use of parentheses you should get a positive radicand whose square root matches (the correct) one of the given choices.