SOLUTION: This questions gives me the jitters... Solve using logs:2^2x+4 minus 2^2x =120 I know the answer but not the steps:(

Question 133315: This questions gives me the jitters...
Solve using logs:2^2x+4 minus 2^2x =120
I know the answer but not the steps:(
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve using logs:2^(2x+4) - 2^2x =120
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Since 2^(a+b) = 2^a*2^b you get:
2^2x*2^4 -2^2x = 120
But 2^4 = 16
16(2^2x) - 2^2x - 120
But 16a - a = 15a.
So you have:
15*2^2x = 120
Divide both sides by 15 to get:
2^2x = 120/15
2^2x = 80
2x log2 = log80
2x = log80/log2
2x = 6.321928..
x = 3.160964...
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Cheers,
Stan H.

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