Help with ogarithms

May 2010
5
0
Well, I'm studying logarithms at the moment, and i have no idea how to do this question :(

Determine the value of x, to 3 d.p.:
3^x = 6^x-1

Help would be much appreciated, and thanks in advance :)
 

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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
Well, I'm studying logarithms at the moment, and i have no idea how to do this question :(

Determine the value of x, to 3 d.p.:
3^x = 6^x-1

Help would be much appreciated, and thanks in advance :)
Do you mean

\(\displaystyle 3^x=6^x-1\)

or

\(\displaystyle 3^x=6^{x-1}\)

?
 

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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
Write

\(\displaystyle log(3^x)=log(6^{x-1})\)

and use the fact that \(\displaystyle log(n^p)=p\cdot log(n)\).
 
May 2010
5
0
Yeah i got that, but i have no idea what to do after that.
 

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MHF Hall of Honor
Mar 2010
2,340
821
Chicago
Yeah i got that, but i have no idea what to do after that.
Well it is straightforward solving for x.

\(\displaystyle log(3^x)=log(6^{x-1})\)

\(\displaystyle x\cdot log(3)=(x-1)\cdot log(6)\)

\(\displaystyle x\cdot log(3)=x\cdot log(6) - log(6)\)

\(\displaystyle x\cdot log(3)-x\cdot log(6)= - log(6)\)

\(\displaystyle x(log(3)-log(6))= - log(6)\)

\(\displaystyle x= \frac{-log(6)}{log(3)-log(6)}\)

\(\displaystyle x= \frac{log(6)}{log(6)-log(3)}\)
 
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