(e^log7 - e^-log7)/(e^log7 + e^-log7) the answer is supposed to be 24/25, but i have no idea where to begin simplifying this (find tanh (log7))
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Originally Posted by twostep08 (e^log7 - e^-log7)/(e^log7 + e^-log7) the answer is supposed to be 24/25, but i have no idea where to begin simplifying this (find tanh (log7)) $\displaystyle e^{\log{7}} = 7$ $\displaystyle e^{-\log{7}} = \frac{1}{7}$ $\displaystyle \frac{7 - \frac{1}{7}}{7 + \frac{1}{7}} = \frac{49-1}{49+1} = \frac{48}{50} = \frac{24}{25} $
ooh.ok.. i dont remember that-what property of logs is that?
Originally Posted by twostep08 ooh.ok.. i dont remember that-what property of logs is that? $\displaystyle b^{\log_b{x}} = x$
And that is, basically, the definition of "logarithm".
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