How do you use the properties of logarithms to expand ${log}_{3} (\frac{\sqrt{a - 1}}{9})$ $log_3(sqrt(a-1)/9)$ ?

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How do you use the properties of logarithms to expand ${log}_{3} (\frac{\sqrt{a - 1}}{9})$ $log_3(sqrt(a-1)/9)$ ?

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Answer 1 · 2017-02-08T02:14:52

$\frac{1}{2} {log}_{3} (a - 1) - 2$ $1/2log_3 (a - 1) - 2$

Explanation:

Use the following laws of logarithms to solve this problem

• ${log a}^{n} = n log a$ $loga^n = nloga$
• $log (\frac{a}{b}) = log a - log b$ $log(a/b) = loga - logb$
$∙ {log}_{a} n = \frac{log n}{log a}$ $•log_an =logn/loga$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$= {log}_{3} \sqrt{a - 1} - {log}_{3} 9$ $=log_3sqrt(a - 1) - log_3 9$

$= {log}_{3} {(a - 1)}^{\frac{1}{2}} - {log}_{3} 9$ $=log_3(a - 1)^(1/2) - log_3 9$

$= \frac{1}{2} {log}_{3} (a - 1) - \frac{log 9}{log 3}$ $= 1/2log_3(a - 1) - log9/log3$

$= \frac{1}{2} {log}_{3} (a - 1) - \frac{{log 3}^{2}}{log 3}$ $= 1/2log_3(a - 1) - log3^2/log3$

$= \frac{1}{2} {log}_{3} (a - 1) - \frac{2 log 3}{log 3}$ $=1/2log_3(a - 1) - (2log3)/log3$

$= \frac{1}{2} {log}_{3} (a - 1) - 2$ $=1/2log_3(a - 1) - 2$

Hopefully this helps!

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How do you use the properties of logarithms to expand ${log}_{3} (\frac{\sqrt{a - 1}}{9})$ $log_3(sqrt(a-1)/9)$ ?

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Explanation:

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How do you use the properties of logarithms to expand ${log}_{3} (\frac{\sqrt{a - 1}}{9})$ $log_3(sqrt(a-1)/9)$ ?