Question 250324: write each expression as a sum and /or difference of logarithms.express powers as factors
1) log [(x+2)/(x+3)^2]
2) In(x.e^x)
3) solve 4^x-2^x-12=0
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! write each expression as a sum and /or difference of logarithms. Express powers as factors
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1) log [(x+2)/(x+3)^2]
Since we are dividing, we can rewrite using subtraction:
log(x+2) - log(x+3)^2
Use the power rule to get:
log(x+2) - 2log(x+3)
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2) In(x.e^x)
I assume you mean
ln(x times e^x)
since we are multiplying we can rewrite using addition
LN(x) + LN(e^x)
using the power rule we get:
LN(x) + x(times)LN(e)
LN(e) = 1, so our answer is:
LN(x) + x
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3) solve 4^x-2^x-12=0
Use a dummy variable. Let Y = 2^x. We get
Y^2 - Y - 12 = 0.
factoring, we get
(Y-4)(Y+3) = 0
Y = 4 and Y = -3.
Resub your 2^x into Y, to get
2^x = 4 and 2^x = -3
2^x = 4 - -> X = 2.
2^x = -3 - -> no solution.
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