SOLUTION: please help me solve: {{{ 2^(2x) - 3 * 2^x - 40= 0 }}}
Algebra.Com
Question 331640: please help me solve:
Found 2 solutions by solver91311, jrfrunner:
Answer by solver91311(24713) (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:
\ =\ \log_2(8))
Apply the laws of logs:
 + \log_b(y) = \log_b(xy))
\ =\ n\log_b(x))
\ =\ 1)
\ =\ \log_2(8))
\ =\ \log_2(2)\ +\ \log_2(2)\ +\ \log_2(2)\ =\ 1\ +\ 1\ +\ 1\ =\ 3)
John
My calculator said it, I believe it, that settles it
Answer by jrfrunner(365) (Show Source): You can put this solution on YOUR website!
Let then
substitute in problem equation
factor this equation
(Y-8)*(y+5)=0
either y=8 or y=-5
-------------------
Case y=8
y = solve for x
ln8/ln2 = x
3=x
----------------
Case y=-5
y =
ln(-5)=x*Ln2 STOP!!!! ln(-5) does not exit, log exist only for values greater than 0
------------------
so there is only one answer: X=3
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