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

$\displaystyle \log_{a} b = c$
$\displaystyle a^c=b$
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

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

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

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

$\displaystyle \log_{2} x = -3$

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

$\displaystyle x=1/8$

and

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

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

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

8/2*2

x=4

i hope i am correct..

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

...

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

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

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

(8/4)*2

x=4

i hope i am correct..