Question 1082044
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I'm going to go way out on a limb and assume that the "9" in the LHS of your equation was really supposed to be an open parenthesis and that what you really meant was:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(-5n\ -\ 1)\ =\ \log_2(20)]


Since you have two equal logs with the same base, then the arguments must be equal, that is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x_1)\ =\ \log_b(x_2)\ \ \Leftrightarrow\ \ x_1\ =\ x_2]


So you have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -5n-1\ =\ 20]


Just solve for *[tex \Large n]


Next time proof-read your post BEFORE you click send so that you save us tutors (who are working for free) the time and trouble of trying to figure out what you mean.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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