# Thread: Logs: Solving for b

1. ## Logs: Solving for b

$-\frac{2}{3} = Log_b\frac{1}{a}$

I need help solving for b.T-T

Thanks a bunch!

2. Originally Posted by hydride
$-\frac{2}{3} = Log_b\frac{1}{a}$

I need help solving for b.T-T

Thanks a bunch!
$-\frac{2}{3} = Log_b\frac{1}{a}$

= $Log_b{a^-1}$

= $-Log_b{a}$

$\frac{2}{3} = Log_b{a}$

$a = b^\frac{2}{3}$

Now find b = ......

3. You'll need this: $\log_{b}(a)\;=\;c\;\iff\;b^{c}\;=\;a$

4. what if c was a negative fraction? i.e $b^{-\frac{2}{3}} = 1/9$

5. Originally Posted by hydride
what if c was a negative fraction? i.e $b^{-\frac{2}{3}} = 1/9$
It should be

$b^{-\frac{2}{3}} = 1/a$

6. oh sorry, i meant 1/9 to begin with.

7. then

$b^{-\frac{2}{3}} = \frac{1}{9}$

$\frac{1}{b^{\frac{2}{3}}} = \frac{1}{9}$

$b^{\frac{2}{3}} = {9}$