How do you use the properties of logarithms to expand $log_3(sqrt(a-1)/9)$ ?

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Answer 1 · 2017-02-08T02:14:52

$1/2log_3 (a - 1) - 2$

Use the following laws of logarithms to solve this problem

• $loga^n = nloga$
• $log(a/b) = loga - logb$
$•log_an =logn/loga$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$=log_3sqrt(a - 1) - log_3 9$

$=log_3(a - 1)^(1/2) - log_3 9$

$= 1/2log_3(a - 1) - log9/log3$

$= 1/2log_3(a - 1) - log3^2/log3$

$=1/2log_3(a - 1) - (2log3)/log3$

$=1/2log_3(a - 1) - 2$

Hopefully this helps!

How do you use the properties of logarithms to expand log_3(sqrt(a-1)/9)log_3(sqrt(a-1)/9)?