Logarithm problem

  • Thread starter helmi
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  • #1
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log2 2x -log3 (3x-1) = 2, solve for x....

you guys dont hv 2 solve the question for me, just guide me to the answer:smile:
 

Answers and Replies

  • #2
arildno
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First step:
Convert one logarithm into its equivalent logarithm expression for the other base.
To do so, the general picture is that for positive number a,b,c, we have:
[tex]c=a^{\log_{a}(c)}=b^{\log_{b}(c)}[/tex]
Taking the a-logarithm of the middle and last expression, we have:
[tex]\log_{a}(c)\log_{a}(a)=\log_{b}(c)\log_{a}(b)[/tex]
That is:
[tex]\log_{b}(c)=\frac{\log_{a}(c)}{\log_{a}(b)}[/tex]
 
  • #3
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ok ive done the 1st step by changing log2 2x to (log3 2x)/(log3 2)

(log3 2x)/(log3 2) - log3 (3x-1) = 2

1.585(log3 2x) -log3 (3x-1) = 2

log3 (2x)^1.585 - log3 (3x-1) = 2

log3 ( (2x^1.585)/(3x-1) ) = 2 , is this correct? then what?
 
  • #4
5
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ok i went ahead & solve the equation to get x=0.522....but when i substitute x=0.522 into the initial equation i did not get 2...what hv i done wrong?
 

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