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

[Acnhduy]

Homework Statement

Solve:

log₅(x-1) + log₅(x-2) - log₅(x+6) = 0

Homework Equations

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.

Please help, thankyou :)

 

[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.

 

[Acnhduy]

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

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

log₅(x-1) + log₅(x-2) - log₅(x+6) = 0

log₅[(x-1)(x-2)] - log₅(x+6) = 0

This gives me

log₅(x²-3x+2) - log₅(x+6) = 0

But

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

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

 

[Mentallic]

What is

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

equivalent to?

 

[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?

 

[Mentallic]

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?

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.

 

[ehild]

Homework Statement

Solve:

log₅(x-1) + log₅(x-2) - log₅(x+6) = 0

Homework Equations

The Attempt at a Solution

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.

Please help, thankyou :)

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

 

[Acnhduy]

Oh I see, thank you!

 

[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?

 

[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.