# Thread: Perhaps a simple log problem here..but Im quite stuck

1. ## Perhaps a simple log problem here..but Im quite stuck

Im quite stuck on starting this problem. I feel totally useles at doing all show that questions, unfortunately they are the hardest question for me and this is no different:

Given that $\displaystyle Log_a x= 2log_a 6 - log_a 3$, show that $\displaystyle x=12$
Do I assume that $\displaystyle log_a$ is $\displaystyle log _{10}$?

Im am actually very stuck, so any help would be greately appreciated kind people

Thanks

2. Originally Posted by fishkeeper
Do I assume that $\displaystyle log_a$ is $\displaystyle log _{10}$?

$\displaystyle \text{\color{red}{no.}}$
$\displaystyle log_a x= 2log_a 6 - log_a 3$

$\displaystyle \Rightarrow log_a x= log_a (6)^2 - log_a 3$

$\displaystyle \Rightarrow log_a x= log_a \frac{(6)^2}{3} = log_a 12$

$\displaystyle \Rightarrow a^{log_a x} = a^{log_a 12}$

$\displaystyle ....$

3. Originally Posted by fishkeeper
Im quite stuck on starting this problem. I feel totally useles at doing all show that questions, unfortunately they are the hardest question for me and this is no different:

Do I assume that $\displaystyle log_a$ is $\displaystyle log _{10}$?

it doesn't make a difference.
this is just an exercise in the properties of logs

$\displaystyle \log{x} = 2\log{6} - \log{3}$

$\displaystyle \log{x} = \log{6^2} - \log{3}$

$\displaystyle \log{x} = \log{36} - \log{3}$

$\displaystyle \log{x} = \log\left(\frac{36}{3}\right)$

$\displaystyle \log{x} = \log{12}$

$\displaystyle x = 12$

4. Thankyou both for taking the time to answer the question.

Now Ive seen the answer in front of me, it seems so simple