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Mathematics 12 Online
OpenStudy (anonymous):

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

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?

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.

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.

OpenStudy (anonymous):

thank you thank you

OpenStudy (ranga):

you are welcome. :)

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