Science Forums 2

CraigD · 09-29-2008

Here’s a proof of
$a^x \overset?= T \left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}$

$\log_a b = \frac{\log a}{\log b}$ , so, for ease of writing, rewrite as
$a^x \overset?= T \left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -1}{\log_Ta -1}\right)}$

Taking the logarithm base a of both sides gives:
$\log_a\left(a^x \right) \overset?= \log_a\left(T \left(\frac{a}{T}\right)^{\left( \frac{x\log_Ta -1}{\log_Ta -1}\right)} \right)$

$\log\left(a^b\right)=b\log a$ , $\log_a a = 1$ , and $\log\left(ab\right)=\log a +\log b$ give
$x \overset?= \log_aT +\frac{x\log_Ta -1}{\log_Ta -1}\log_a\left(\frac{a}{T}\right)$

And
$x \overset?= \log_aT +\frac{x\log_Ta -1}{\log_Ta -1} \left(1 -\log_aT \right)$

Multiply both sides by $\log_Ta -1$ giving
$(\log_Ta -1)x \overset?= (\log_Ta -1)\log_aT +(x\log_Ta -1)(1 -\log_aT)$

Expand, giving
$x\log_Ta -x \overset?= \log_aT\log_Ta -\log_aT +x\log_Ta -1 -x\log_aT\log_Ta +\log_aT$

$\log_ab\log_ba = 1$ , so
$x\log_Ta -x \overset?= 1 -\log_aT +x\log_Ta -1 -x +\log_aT$

And simplify, giving
$-x = -x$

Science Forums 2

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