SOLUTION: How can I solve this problem algebraically ln(x + 3) � ln(x � 4) = ln(x)

Question 859248: How can I solve this problem algebraically ln(x + 3) � ln(x � 4) = ln(x)
Found 2 solutions by jim_thompson5910, rothauserc:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=-5, and c=-3

Negate -5 to get 5

Square -5 to get 25 (note: remember when you square -5, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Now break up the fraction

or

Simplify

or

So the solutions are:

or

Those are the possible solutions. You need to check each one. I'll let you do this part.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
ln(x + 3) � ln(x � 4) = ln(x)
ln [ (x+3) / (x-4) ] = ln(x)
ln [ (x+3) / (x-4) ] - ln(x) = 0
ln [ ((x+3) / (x-4)) / x ] = 0
ln [ (x+3) / x(x-4) ] = 0
now x cannot be less than or equal to 4 since ln is not defined for 0 or negative values,
therefore x must be greater than 4

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