If $log_a 36=2.2265$ and $log_a 4=.8614$ , how do you evaluate $log_a 9$ ?

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Answer 1 · 2015-06-11T17:51:58

Using the property of $log$ which states that $log_n(a/b)=log_na-log_nb$

Following the $log$ property, we can use your pieces of information as follows:

$log9=log(36/4)$

Thus,

$log_a36-log_a4=log_a9$

$2.2265-0.8614=log_a9$

$log_a9=1.3651$

Following $log$ definition, we have that $log_ab=c <=> a^c=b$

Thus,

$a^(1.3651)=9$

To solve this, isolating $a$ we can follow one property of exponentials, which states $(a^n)^m=a^(n*m)$ by elevating both sides to $(1/1.3651)$ :

$(a^(cancel(1.3651)))^(1/cancel(1.3651))=9^(1/1.3651)$

$a=9^(1/1.3651)$

$a~=0.7325$

If log_a 36=2.2265log_a 36=2.2265 and log_a 4=.8614log_a 4=.8614, how do you evaluate log_a 9log_a 9?