Question 331640
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^{2x}\ -\ 3\cdot2^x\ -\ 40\ =\ 0]


Eww...that is uglier than a mud fence.  What do you do with an ugly face?  Put a little make-up on it.


Let *[tex \Large u\ =\ 2^x] and then substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 3u\ -\ 40\ =\ 0]


Now that's something familiar we can deal with.  Let's see...-8 times 5 is -40 and -8 plus 5 is -3, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 8]


or 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ -5]


Substitute back:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^x\ =\ 8\ \ \ \Rightarrow\ \ x\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^x\ =\ -5]


Stop right there.  *[tex \LARGE -5] is NOT in the range of *[tex \LARGE 2^x].  Discard the extraneous root.


Solution set is *[tex \Large \{3\}]


If this step:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^x\ =\ 8\ \ \ \Rightarrow\ \ x\ =\ 3]


confuses you, try it this way.  Take the base 2 log of both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_2(2^x)\ =\ \log_2(8)]


Apply the laws of logs:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \log_b(x) + \log_b(y) = \log_b(xy)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \log_b(b)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\log_2(2)\ =\ \log_2(8)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \log_2(2\cdot2\cdot2)\ =\ \log_2(2)\ +\ \log_2(2)\ +\ \log_2(2)\ =\ 1\ +\ 1\ +\ 1\ =\ 3]



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">
</font>