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How do you expand this logarithm?

${log}_{3} (z^{4} \sqrt{x})$ $log_(3)(z^(4)sqrtx)$

2 Answers

Nov 21, 2016

$4 {log}_{3} (z) + \frac{1}{2} {log}_{3} (x)$ $4log_3 (z) + 1/2log_3(x)$

Explanation:

General Rules:
$XXX {log}_{b} a^{c} = c \cdot {log}_{b} a$ $color(white)("XXX")log_b a^c = c * log_b a$

$XXX {log}_{b} (a \cdot c) = {log}_{b} + {log}_{b} c$ $color(white)("XXX")log_b (a * c) = log_b + log_b c$

Nov 21, 2016

$4 {log}_{3} (z) + \frac{1}{2} {log}_{3} (x)$ $4log_3(z)+1/2log_3(x)$ .

Explanation:

By using the rule of logarithms where $log (a \cdot b) = log a + log b$ $log(a*b)=loga+logb$ , we first get

${log}_{3} (z^{4} \sqrt{x})$ $log_3(z^4sqrtx)$
$= {log}_{3} (z^{4}) + {log}_{3} (\sqrt{x})$ $=log_3(z^4)+log_3(sqrtx)$
$= {log}_{3} (z^{4}) + {log}_{3} (x^{1 / 2})$ $=log_3(z^4)+log_3(x^(1//2))$

Another rule of logarithms is $log (a^{b}) = b log (a)$ $log(a^b)=blog(a)$ . We now use this to get

$= 4 {log}_{3} (z) + \frac{1}{2} {log}_{3} (x)$ $=4log_3(z)+1/2log_3(x)$

Unless you have been asked to rewrite the "base 3" logarithms in "base 10" form, this is as much expansion as we can do.

Hope this helps!

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