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- Jul 20th 2006, 07:40 AMLane2 Questions!!
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- Jul 20th 2006, 08:03 AMCaptainBlackQuote:

Originally Posted by**Lane**

Divide through by $\displaystyle 3$ to get:

$\displaystyle e^{4x-5}=2/3$

Now take natural logs of both sides:

$\displaystyle

4x-5=\log_e(2/3)

$.

Add $\displaystyle 5$ to both sides:

$\displaystyle

4x=\log_e(2/3)+5

$,

finally divide through by $\displaystyle 4$:

$\displaystyle

x=(\log_e(2/3)+5)/4 = \log_e \left(\root {1/4}\of {2/3} \right) + 5/4

$

RonL - Jul 20th 2006, 08:09 AMCaptainBlackQuote:

Originally Posted by**Lane**

Now

$\displaystyle

(f \circ g)(x)=f(g(x))=4 g(x)-5=$$\displaystyle 4(x^2-2x+6)-5=4x^2-8x+19

$

RonL