e^0.3=2
Someone give me step by step please
I get there are rules I don't know how to use them just talk me through it
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OpenStudy (ranga):
There is something missing in the problem. Is there an x or some other variable somewhere?
OpenStudy (anonymous):
oh yes im sorry e^0.3x=2
x is part of the exponent
OpenStudy (ranga):
e^(0.3x) = 2
To bring the x down from the exponent, you take natural logarithm on both sides.
ln(e^(0.3x)) = ln(2)
(0.3x) * ln(e) = ln(2)
0.3x * 1 = ln(2)
x = ln(2) / 0.3 = ?
OpenStudy (ranga):
\[\Large \ln(A ^{n}) = n * \ln(A)\]
OpenStudy (anonymous):
2.31?
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OpenStudy (ranga):
yes.
OpenStudy (anonymous):
awesome sauce!
OpenStudy (ranga):
alright. Do you have any questions about any of the steps or is it pretty clear?
OpenStudy (anonymous):
take me through one more problem? maybe that will help?
OpenStudy (ranga):
ok. go ahead and post the problem.
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OpenStudy (anonymous):
3=log(2x-1) I got x+500.5 I'm
just am not sure if this is right
OpenStudy (anonymous):
I mean x=500.5
OpenStudy (ranga):
3=log(2x-1) or log(2x-1) = 3
means: 2x - 1 = 10^3 = 1000
2x = 1001
x = 500.5
yes, you have the correct answer.