Question 94675: KEEP GETTING THESE WRONG - HELP PLEASE
Need to solve:
log9(1/3)=y
Need to simplify:
e^x * 4e^2x+1 (the 2x+1 is all in exponent)
Thanks
LLO Found 2 solutions by Nate, stanbon:Answer by Nate(3500) (Show Source):
You can put this solution on YOUR website! log9(1/3)=y
log9(3^-1)=y
-log9(3)=y
3=9^-y
-log10(3)/log10(9)=y
-log10(3)/log10(3^2)=y
-log10(3)/[2*log10(3)]=y
-1/2=y
~~~~~~~~~~~~~~
e^x * 4e^(2x + 1)
4(e^x)e^(2x + 1)
4e^(x + 2x + 1)
4e^(3x + 1)
You can put this solution on YOUR website! Need to solve:
log9(1/3)=y
Write in exponential form:
9^y = 1/3
3^2y = 3^-1
2y=-1
y = -1/2
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Need to simplify:
e^x * 4e^(2x+1)
= 4 e^(2x+1+x)
= 4^(3x+1)
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Cheers,
Stan H>
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