5e^.2x=7

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- Dec 21st 2009, 10:26 AMmath321logarithmic problem
5e^.2x=7

- Dec 21st 2009, 10:36 AMbigwave
first just need to clarify do you mean

$\displaystyle 5 e^{.2}x=7$ or $\displaystyle 5e^{.2x}=7 $ - Dec 21st 2009, 10:38 AMtonio
- Dec 21st 2009, 10:42 AMmath321
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 AMbigwave
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 AMmath321
thanks alot

i got an exam tomoro so i am just practicing