Logs: Solving for b

Sep 2009
51
0
\(\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}\)

I need help solving for b.T-T

Thanks a bunch!
 
Jun 2009
806
275
\(\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}\)

I need help solving for b.T-T

Thanks a bunch!
\(\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}\)

= \(\displaystyle Log_b{a^-1}\)

= \(\displaystyle -Log_b{a}\)

\(\displaystyle \frac{2}{3} = Log_b{a}\)

\(\displaystyle a = b^\frac{2}{3}\)

Now find b = ......
 
Aug 2007
3,171
860
USA
You'll need this: \(\displaystyle \log_{b}(a)\;=\;c\;\iff\;b^{c}\;=\;a\)
 
Sep 2009
51
0
what if c was a negative fraction? i.e \(\displaystyle b^{-\frac{2}{3}} = 1/9\)
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
then

\(\displaystyle b^{-\frac{2}{3}} = \frac{1}{9}\)

\(\displaystyle \frac{1}{b^{\frac{2}{3}}} = \frac{1}{9}\)

\(\displaystyle b^{\frac{2}{3}} = {9}\)