What is ${log}_{4} (2 \sqrt{2})$ $log_4(2sqrt(2))$ ?

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What is ${log}_{4} (2 \sqrt{2})$ $log_4(2sqrt(2))$ ?

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Answer 1 · 2016-03-26T00:32:32

${log}_{4} (2 \sqrt{2}) = \frac{3}{4}$ $log_4 (2sqrt(2)) = 3/4$

Explanation:

The change of base formula tells us that if $a, b, c > 0$ $a, b, c > 0$ and $a, c \neq 1$ $a, c != 1$ then:

${log}_{a} b = \frac{{log}_{c} b}{{log}_{c} a}$ $log_a b = (log_c b) / (log_c a)$

Hence:

${log}_{4} (2 \sqrt{2}) = \frac{{log}_{2} (2 \sqrt{2})}{{log}_{2} (4)} = \frac{{log}_{2} (2^{\frac{3}{2}})}{{log}_{2} (2^{2})} = \frac{\frac{3}{2}}{2} = \frac{3}{4}$ $log_4 (2sqrt(2)) = log_2 (2sqrt(2)) / log_2 (4) = log_2 (2^(3/2)) / log_2 (2^2) = (3/2) / 2 = 3/4$

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What is ${log}_{4} (2 \sqrt{2})$ $log_4(2sqrt(2))$ ?

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