Solve Logarithm Problem: log2 2x -log3 (3x-1) = 2, Find X

[helmi]

log2 2x -log3 (3x-1) = 2, solve for x...

you guys don't hv 2 solve the question for me, just guide me to the answer

 

[arildno]

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:
$$c=a^{\log_{a}(c)}=b^{\log_{b}(c)}$$
Taking the a-logarithm of the middle and last expression, we have:
$$\log_{a}(c)\log_{a}(a)=\log_{b}(c)\log_{a}(b)$$
That is:
$$\log_{b}(c)=\frac{\log_{a}(c)}{\log_{a}(b)}$$

 

[helmi]

ok I've 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?

 

[helmi]

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?