# logarithmic problem

• Dec 21st 2009, 10:26 AM
math321
logarithmic problem
5e^.2x=7
• Dec 21st 2009, 10:36 AM
bigwave
first just need to clarify do you mean

$\displaystyle 5 e^{.2}x=7$ or $\displaystyle 5e^{.2x}=7$
• Dec 21st 2009, 10:38 AM
tonio
Quote:

Originally Posted by math321
5e^.2x=7

Hint: if $\displaystyle e^b=y$ , then taking natural logarithm on both sides we get $\displaystyle b=\ln y$ , provided, of course, $\displaystyle y>0$

Tonio
• Dec 21st 2009, 10:42 AM
math321
the second sum

and i want to find out...instead of I using the up arrow how do i get someting raised to a power
• Dec 21st 2009, 10:57 AM
bigwave
ln(5e^.2x) = ln7

is

$\displaystyle \frac{x}{5} + ln5 = ln7$

so $\displaystyle x = 5(ln7 - ln5)$ or $\displaystyle 5ln\frac{7}{5}$

use latex to get the power

x^2 then highlight the expression with $\displaystyle \Sigma$icon and you get $\displaystyle x^2$ after it is compiled in preview or after saving and posting
• Dec 21st 2009, 11:06 AM
math321
thanks alot

i got an exam tomoro so i am just practicing