# Math Help - Logarithms..(solve)

1. ## Logarithms..(solve)

Hi I cannot wrap my head around these two problems.
Some help would be greatly appreciated.

and

2. ## log laws

these questions are a matter of applying log laws

$\log_{a} b = c$
$a^c=b$
so for your case a=2 and c=-3, so sub them into the second equation

the second question is a combination of log laws

$e \log_{a} b = \log_{a} b^e$

and $\log_{a} b + \log_{a} d = \log_{a} (b*d)$
and $\log_{a} b - \log_{a} d = \log_{a} (\frac{b}{d})$
so for the second question rewrite the equation in the form of $\log_{a} b^e$ and then apply the addition and subtraction rules to get your answer

3. thank you, i think i got it..

$\log_{2} x = -3$

$2^{-3}=x$

$x=1/8$

and

${3/2} \log_{b} 4 - {2/3} \log_{b}8 + \log_{b}2= \log_{b} x$

$\log_{b}4^{3/2}-\log_{b}8^{2/3} + \log_{b}2= \log_{b}x$

$(4^{3/2}) / (8^{2/3})*2=x$

8/2*2

x=4

i hope i am correct..

4. Originally Posted by Macabro
thank you, i think i got it..

...

${3/2} \log_{b} 4 - {2/3} \log_{b}8 + \log_{b}2= \log_{b} x$

$\log_{b}4^{3/2}-\log_{b}8^{2/3} + \log_{b}2= \log_{b}x$

$(4^{3/2}) / (8^{2/3})*2=x$

(8/4)*2

x=4

i hope i am correct..