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Mathematics 17 Online
OpenStudy (anonymous):

Logarithm NO DECIMALS ln 12x^2 - ln3x

OpenStudy (ranga):

You can make use of the following ientities to solve the problem: \[\ln(AB) = \ln(A) + \ln(B)\]\[\ln(A/B) = \ln(A) - \ln(B)\]\[\ln(x ^{n}) = nln(x)\]

OpenStudy (ranga):

*identities*

OpenStudy (ranga):

\[\Large \ln(12x ^{2}) -\ln(3x) = \ln(\frac{ 12x^{2} }{ 3x }) = \ln(4x)\]

OpenStudy (anonymous):

so is that the final?

OpenStudy (ranga):

If you want you can write ln(4x) as ln(4) + ln(x). But it is up to you. Either one should be okay.

OpenStudy (anonymous):

can you do another one for me

OpenStudy (ranga):

sure.

OpenStudy (anonymous):

e^x-1 - 3 =2

OpenStudy (ranga):

Is the exponent just x or (x-1)?

OpenStudy (anonymous):

\[e ^{x-1} -3 =2\]

OpenStudy (ranga):

OK. Isolate the x term on the left. So add 3 to both sides: e^(x-1) = 5 Take natural logarithm on both sides. On the left ln and e will cancel out leaving just (x-1) (x-1) = ln5 add 1 to both sides: x = 1 + ln5

OpenStudy (anonymous):

is that the final

OpenStudy (ranga):

yes.

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