# 2 Questions!!

• July 20th 2006, 07:40 AM
Lane
2 Questions!!
questions are on attachment!
• July 20th 2006, 08:03 AM
CaptainBlack
Quote:

Originally Posted by Lane
questions are on attachment!

1. Solve $3\ e^{4x-5}=2$

Divide through by $3$ to get:

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

Now take natural logs of both sides:

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

Add $5$ to both sides:

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

finally divide through by $4$:

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

RonL
• July 20th 2006, 08:09 AM
CaptainBlack
Quote:

Originally Posted by Lane
questions are on attachment!

2. Find $f\circ g$, where $f(x)=4x-5$, and $g(x)=x^2-2x+6$.

Now

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

RonL