Solve Log Equation for x: 2log3x + log3(x-1) = log32 + log36x | Homework Help

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To solve the equation 2log3x + log3(x-1) = log32 + log36x, the initial step leads to log3(x^3 - x^2) = log312x. Cancelling the logs results in the polynomial x^3 - x^2 = 12x, which can be rearranged to x(x^2 - x - 12) = 0. Factoring reveals one solution as x=0, but this is extraneous since log30 is undefined. The remaining solutions come from solving the quadratic equation x^2 - x - 12 = 0.
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Homework Statement


Solve for x:
2log3x + log3(x-1) = log32 + log36x

Homework Equations


N/A


The Attempt at a Solution


Ok, so I tried to solve it and I came up with this and got stuck.

log3(x3-x2) = log312x
Even if I canceled out the logs it would be x3-x2 = 12x and then how would I solve?
 
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Welcome to PF!

Hi chops369! Welcome to PF! :smile:
chops369 said:
Solve for x:
2log3x + log3(x-1) = log32 + log36x

log3(x3-x2) = log312x
Even if I canceled out the logs it would be x3-x2 = 12x and then how would I solve?

Divide both sides by x. :wink:
 
Ohhhh, I see now. I didn't know you had to factor. Thanks a lot.
 
Careful of dividing by x! It's OK in this problem since x=0 is an extraneous solution, but you do run the risk of losing a solution.
 
It is better if you factor x:
x3-x2-12x=0

x(x2-x-12)=0

One of the solutions is x1=0

The other two come from the quadratic equation.

As you can see zero, can't be solution because if you put the 0 in the first equation, it would not be possible since log30 doesn't exist.

Regards.
 
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