2
$\begingroup$

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.

$\endgroup$
1
  • 6
    $\begingroup$ Note that $(\log8)^2\ne\log(8^2)$. $\endgroup$
    – Kenta S
    Commented Mar 7, 2021 at 20:35

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .