Mathematics
15 Online
OpenStudy (saifoo.khan):
log
9
x = a log
3
x
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OpenStudy (anonymous):
can you type this in the equation editor?
OpenStudy (saifoo.khan):
Find a
OpenStudy (anonymous):
im not sure how to read what you posted
OpenStudy (saifoo.khan):
\[\log_{9} x = a \log_{3}x \]
OpenStudy (saifoo.khan):
find a
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OpenStudy (saifoo.khan):
ans?
OpenStudy (anonymous):
i get 1/2
OpenStudy (anonymous):
do you know change of base formula?
OpenStudy (saifoo.khan):
correct. but how?
OpenStudy (saifoo.khan):
umm, i knw half of it.
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OpenStudy (anonymous):
\[\log_{a} b=\frac{\ln(b)}{\ln(a)}\]
OpenStudy (anonymous):
change of base formula: \[\log_{a} x=\log_{b} x/\log_{b} a \]
OpenStudy (anonymous):
you can use that formula to cancel out the terms with x and solve for a
OpenStudy (saifoo.khan):
can u plz giv me its solution too? i will b very thnkful. :s
OpenStudy (anonymous):
\[\frac{\ln(x)}{\ln(9)}=a \frac{\ln(x)}{\ln(3)} \rightarrow \frac{\ln(3)}{\ln(9)}=a\]
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OpenStudy (anonymous):
\[\frac{\ln(3)}{\ln(3^2)}=a \rightarrow \frac{\ln(3)}{2\ln(3)}=a \rightarrow \frac{1}{2}=a\]
OpenStudy (anonymous):
ye, rsviale is correct
OpenStudy (saifoo.khan):
oh! now i got it, thnks guyz! :D
OpenStudy (saifoo.khan):
i shall try the 2nd 1.. if u cant find its soultion thn u guyz hav to help!
OpenStudy (anonymous):
k ill try to help if you need it
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OpenStudy (saifoo.khan):
how u took\[\log_{} 3 \div 2\log_{} 3\]
OpenStudy (saifoo.khan):
how u took log 3 in the numerator??