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Mathematics 10 Online
OpenStudy (anonymous):

Log[3]40-log[3]10

OpenStudy (anonymous):

Would the answer be log[3]4

OpenStudy (anonymous):

\[\log_{}a - \log_{}b = \log_{} \frac{ a }{ b }\]

OpenStudy (anonymous):

Yup..u r Correct..:)

OpenStudy (anonymous):

1/16=64^4x-3

OpenStudy (anonymous):

X=7/12?

OpenStudy (anonymous):

\[2^{-4}=2^{6(4x-3)}\] \[6(4x-3)=-4\] etc

OpenStudy (anonymous):

24x-18=-4

OpenStudy (anonymous):

you are right, it is \(\frac{7}{12}\)

OpenStudy (anonymous):

3 log 2x=4

OpenStudy (anonymous):

log base ten?

OpenStudy (anonymous):

X=10.772

OpenStudy (anonymous):

\[\log(2x)=\frac{4}{3}\] \[2x=10^{\frac{4}{3}}\] \[x=\frac{1}{2}10^{\frac{4}{3}}\]

OpenStudy (anonymous):

10.772 is correct rounded

OpenStudy (anonymous):

Log 2x=4/3 X=104/3 divided by 2

OpenStudy (anonymous):

Ln(5x+7)=8

OpenStudy (anonymous):

X=(e^8-7)/5 X=594.792

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

5e^2x + 11 = 30 X=4.604

OpenStudy (anonymous):

That was the last one:)

OpenStudy (anonymous):

\[2x=\ln(\frac{19}{5})\] \[x=\frac{1}{2}\ln(\frac{19}{5})\]

OpenStudy (anonymous):

i get a different answer for the last one

OpenStudy (anonymous):

but maybe i read it wrong is it \[5e^{2x}+11\] ir \[5e^{2x+11}\]3log(2x)=4

OpenStudy (anonymous):

i mean \[5e^{2x}+11\] or \[5e^{2x+11}\]

OpenStudy (anonymous):

Second one.

OpenStudy (anonymous):

then you are off by a minus sign, should be \(-4.6\)

OpenStudy (anonymous):

or \(-4.604\)

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