My question has to do with simplifying.
$\displaystyle e^(-9ln(x)) = 1/e^(9ln(x)) $
What does this simplify to?
[Math] 1/9x [/tex] ?
A basic property of logs leads to the conclusion that, for any base "b", the following holds:
. . . . .$\displaystyle b^{\log_b(m)}\, =\, m$
Applying a log rule inside that power to put that -9 back inside the log, we get a power of:
. . . . .$\displaystyle -9\ln(x)\, =\, \ln(x^{-9})\, =\, \ln\left(\frac{1}{x^9}\right)$
...by the rule for negative exponents.
Now apply the first rule above to simply the exponential expression.![]()