$\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}$ I need help solving for b.T-T Thanks a bunch!
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Originally Posted by hydride $\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}$ 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 = ......
You'll need this: $\displaystyle \log_{b}(a)\;=\;c\;\iff\;b^{c}\;=\;a$
what if c was a negative fraction? i.e $\displaystyle b^{-\frac{2}{3}} = 1/9$
Originally Posted by hydride what if c was a negative fraction? i.e $\displaystyle b^{-\frac{2}{3}} = 1/9$ It should be $\displaystyle b^{-\frac{2}{3}} = 1/a$
oh sorry, i meant 1/9 to begin with.
then $\displaystyle b^{-\frac{2}{3}} = \frac{1}{9}$ $\displaystyle \frac{1}{b^{\frac{2}{3}}} = \frac{1}{9}$ $\displaystyle b^{\frac{2}{3}} = {9}$
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