# Solve log5(x-1) + log5(x-2) - log5(x+6) = 0

1. Oct 3, 2013

### Acnhduy

1. The problem statement, all variables and given/known data

Solve:

log5(x-1) + log5(x-2) - log5(x+6) = 0

2. Relevant equations

3. The attempt at a solution

I feel like this is really simple but I cannot get the answer.

This is what I attempted

Since they have the same base, I multiplied the top

log5[(x-1)(x-2)] - log5(x+6) = 0

This gives me

log5(x2-3x+2) - log5(x+6) = 0

At this point I did not now what to do, I tried 2 methods,

1.
Since they are the same base again, I divided the top this time

log5(x2-3x+2) / (x+6) = 0

but I don't know how to continue

2.

I moved - log5(x+6) to the other side

log5(x2-3x+2) = log5(x+6)

Since they have the same base I ignored them so

x2-3x+2 = x+6

Which gave me

x2 - 4x - 4

Which cannot be factored.

Last edited: Oct 4, 2013
2. Oct 4, 2013

### Mentallic

What you've done is incorrect because the rule of logarithms is

$$\log{a}+\log{b} = \log{ab}$$

But

$$a^m\cdot a^n = a^{m+n}\neq a^{mn}$$

See if you can apply these algebraic formulae to correctly answer your question.

3. Oct 4, 2013

### Acnhduy

I did use this logarithm law, since the base for all of them are common.

But

I did not use at all throughout. I am not sure where I have gone wrong, can you point it out for me?

Last edited: Oct 4, 2013
4. Oct 4, 2013

### Mentallic

What is

$$\log{a^m}+\log{a^n}$$

equivalent to?

5. Oct 4, 2013

### Acnhduy

Oh, I think I found the problem... 5 is the base, not 10.

So this is the correct equation:

log5(x-1) + log5(x-2) - log5(x+6) = 0

Terribly sorry, would my calculations be correct following the equation above?

6. Oct 4, 2013

### Mentallic

Oh, in that case, you're on the right track in #2 in your first post. $x^2-4x-4=0$ cannot be factored in the manner you're thinking of, but it does have irrational roots. Use the quadratic formula to find them.

Also, keep in mind that some (or even all) of your x solutions might not be valid. You need to plug them back into the original equation to see if it makes sense. If you end up with log(-1) for example, then that solution of x needs to be scrapped.

7. Oct 4, 2013

### ehild

What is x2 - 4x - 4 equal to?

It should be a quadratic equation, can you solve it? The roots need not be integer numbers!

ehild

8. Oct 4, 2013

### Acnhduy

Oh I see, thank you!

9. Oct 4, 2013

### Acnhduy

x2 - 4x - 4 = 0

When you mean that they do not need to be intergers, how do I go about solving? As suggested by Mentallic, I would have to use the quadratic formula right?

10. Oct 4, 2013

### Mentallic

The quadratic formula is the easiest method, but completing the square works too if you don't remember the formula.

$$x^2-4x-4=0$$

Now add 4 to both sides.
$$(x^2-4x+4)-4=4$$

$$(x-2)^2-4=4$$

etc.

This is how the quadratic formula is derived. Begin with

$$ax^2+bx+c=0$$

and then complete the square to solve for x.