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Mathematics 15 Online
OpenStudy (saifoo.khan):

log 9 x = a log 3 x

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

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.

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):

\[\frac{\ln(x)}{\ln(9)}=a \frac{\ln(x)}{\ln(3)} \rightarrow \frac{\ln(3)}{\ln(9)}=a\]

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

OpenStudy (saifoo.khan):

how u took\[\log_{} 3 \div 2\log_{} 3\]

OpenStudy (saifoo.khan):

how u took log 3 in the numerator??

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