solve the equation log(x-5) + log(x-2) = 1 algebraically. list any extraneous solutions and explain. if there are no extraneous solutions explain why not?

## list any extraneous solutions and explain.

# 2 Answers

When you're solving a logarithmic equation, the first thing you want to do is get all of the terms with a log on one side and move the constant terms to the other side. Since that's already taken care of, now you want to combine the log terms into one logarithm.
The sum of 2 logs is the log of the product, so we end up with:

log((x-5)(x-2))=1, or log(x^{2 }-7x+10) = 1.

Now we can convert the logarithmic equation into an exponential equation. The base of the logarithm is 10 since it's not otherwise specified, so we now have:

10^{1 }= x^{2 }-7x+10, or 10 = x^{2 }- 7x +10.

From there, we just have to solve this quadratic equation. We first subtract 10 from both sides, leaving us:

x^{2} - 7x = 0.

If we factor the left side, we are left with: x(x-7)=0. Solving this, we get x=0 or x=7.

Now, we have to check if either of the solutions is extraneous. We'll start with x=0.

If x=0, then log(0-5) + log(0-2) = 1, giving us log(-5) + log(-2) = 1, which is impossible since a logarithm cannot have a negative argument. Therefore, x=0 is an extraneous solution.

If x=7, then log(7-5) + log(7-2) = 1, giving us log(2) + log(5) = 1. Using the previous property that the sum of two logs is the product, then:

log (2*5)=1

log (10) =1

10^{1} = 10, which checks out!

So x=7 is the solution, and x=0 is extraneous.

log(x-5) + log(x-2) = 1

log((x-5)(x-2)) = 1

(x-5)(x-2) = 10^1

(x-5)(x-2) = 10

x^2-7x+10 = 10

x^2-7x = 0

x(x-7) = 0

x = {0, 7}

0 is an "extraneous" solution because you cannot plug it back into the original equation.

Solution: x = 7