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)
$$ (\log{2^3})^2 + (\log{2^{-4}})^2 $$ $$ (3\log{2})^2 + (-4 \log{2})^2 $$ $$ 9 (\log{2})^2 + 16 (\log{2})^2 $$ $$ 25 (\log{2})^2 $$
$$ \sqrt{25 (\log{2})^2} $$ $$ 5 \log{2} $$
Answer is D.
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.