SOLUTION: please help me solve: {{{ 2^(2x) - 3 * 2^x - 40= 0 }}}

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Question 331640: please help me solve: +2%5E%282x%29+-+3+%2A+2%5Ex+-+40=+0+
Found 2 solutions by solver91311, jrfrunner:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




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

Let and then substitute:



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



or



Substitute back:





Stop right there. is NOT in the range of . Discard the extraneous root.

Solution set is

If this step:



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



Apply the laws of logs:












John

My calculator said it, I believe it, that settles it

Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
+2%5E%282x%29+-+3+%2A+2%5Ex+-+40=+0+
Let y=2%5Ex then Y%5E2=%282%5Ex%29%5E2+=+2%5E%282x%29
substitute in problem equation
Y%5E2+-+3%2Ay+-40+=0
factor this equation
(Y-8)*(y+5)=0
either y=8 or y=-5
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Case y=8
y = 8+=+2%5Ex solve for x
ln8+=+ln%282%5Ex%29+=+x%2Aln2
ln8/ln2 = x
3=x
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Case y=-5
y = -5+=+2%5Ex
ln(-5)=x*Ln2 STOP!!!! ln(-5) does not exit, log exist only for values greater than 0
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so there is only one answer: X=3