Mathematics
10 Online
OpenStudy (anonymous):
Log[3]40-log[3]10
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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?
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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?
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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
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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):
\[2x=\ln(\frac{19}{5})\]
\[x=\frac{1}{2}\ln(\frac{19}{5})\]
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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 \(-4.6\)
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OpenStudy (anonymous):
or \(-4.604\)