H hydride Sep 2009 51 0 May 12, 2010 #1 \(\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}\) I need help solving for b.T-T Thanks a bunch!

S sa-ri-ga-ma Jun 2009 806 275 May 12, 2010 #2 hydride said: \(\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}\) I need help solving for b.T-T Thanks a bunch! Click to expand... \(\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}\) = \(\displaystyle Log_b{a^-1}\) = \(\displaystyle -Log_b{a}\) \(\displaystyle \frac{2}{3} = Log_b{a}\) \(\displaystyle a = b^\frac{2}{3}\) Now find b = ......

hydride said: \(\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}\) I need help solving for b.T-T Thanks a bunch! Click to expand... \(\displaystyle -\frac{2}{3} = Log_b\frac{1}{a}\) = \(\displaystyle Log_b{a^-1}\) = \(\displaystyle -Log_b{a}\) \(\displaystyle \frac{2}{3} = Log_b{a}\) \(\displaystyle a = b^\frac{2}{3}\) Now find b = ......

T TKHunny Aug 2007 3,171 860 USA May 12, 2010 #3 You'll need this: \(\displaystyle \log_{b}(a)\;=\;c\;\iff\;b^{c}\;=\;a\)

H hydride Sep 2009 51 0 May 12, 2010 #4 what if c was a negative fraction? i.e \(\displaystyle b^{-\frac{2}{3}} = 1/9\)

S sa-ri-ga-ma Jun 2009 806 275 May 12, 2010 #5 hydride said: what if c was a negative fraction? i.e \(\displaystyle b^{-\frac{2}{3}} = 1/9\) Click to expand... It should be \(\displaystyle b^{-\frac{2}{3}} = 1/a\)

hydride said: what if c was a negative fraction? i.e \(\displaystyle b^{-\frac{2}{3}} = 1/9\) Click to expand... It should be \(\displaystyle b^{-\frac{2}{3}} = 1/a\)

pickslides MHF Helper Sep 2008 5,237 1,625 Melbourne May 12, 2010 #7 then \(\displaystyle b^{-\frac{2}{3}} = \frac{1}{9}\) \(\displaystyle \frac{1}{b^{\frac{2}{3}}} = \frac{1}{9}\) \(\displaystyle b^{\frac{2}{3}} = {9}\)

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