# Math Help - Math 12 logarithms

1. ## Math 12 logarithms

Hi all,

I just a few questions on Math 12, not sure how to do these questions (I need to know the how-to more than the answer itself):

1) "The population of a type of bacteria triples every 20 hours. In how many hours will a population of 30 become a population of 1000?"

2) "Solve for x: log(subcript a) a ^ 2x = log (subscript b^2) b^(3x-3)

3) Simplify: [log (subscript: 1/x) 1/y ] - [log (subscript 1/x) y ] - [log (subscript X) 1/y]

(This is from the Logarithms unit).

Thanks a lot in advance!

2. Originally Posted by funnytim
Hi all,

I just a few questions on Math 12, not sure how to do these questions (I need to know the how-to more than the answer itself):

1) "The population of a type of bacteria triples every 20 hours. In how many hours will a population of 30 become a population of 1000?"

2) "Solve for x: log(subcript a) a ^ 2x = log (subscript b^2) b^(3x-3)

3) Simplify: [log (subscript: 1/x) 1/y ] - [log (subscript 1/x) y ] - [log (subscript X) 1/y]

(This is from the Logarithms unit).

Thanks a lot in advance!
2) $\log_a(a^{2x})=log_b(b^{3x-3})$.

Do you remember the rule $\log_a(a^m)=m$?

So we get $2x=3x-3$. Can you solve for x?

3) $\log_{\frac{1}{x}}(\frac{1}{y})-\log_{\frac{1}{x}}(y)-\log_x(\frac{1}{y})$

Do you remember the rule $\log_a(m)-\log_a(n) = \log_a(\frac{m}{n})$? Using this we get

$\log_{\frac{1}{x}}(\frac{\frac{1}{y}}{y})-\log_x(\frac{1}{y})$
$=\log_{\frac{1}{x}}(\frac{1}{y^2})-\log_x(\frac{1}{y})$

We can't do any more as the bases of the logarithms are not the same.

3. Hold on there, Pardner.

$log_{1/x}(a) = \frac{log(a)}{log(1/x)} = \frac{log(a)}{-log(x)} = -log_{x}(a)$

4. This looks about right:

5. Hey, thanks guys!