• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter Acnhduy
  • Start date
  • #1
31
1

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 :)
 
Last edited:

Answers and Replies

  • #2
Mentallic
Homework Helper
3,798
94
What you've done is incorrect because the rule of logarithms is

[tex]\log{a}+\log{b} = \log{ab}[/tex]

But

[tex]a^m\cdot a^n = a^{m+n}\neq a^{mn}[/tex]

See if you can apply these algebraic formulae to correctly answer your question.
 
  • #3
31
1
[tex]\log{a}+\log{b} = \log{ab}[/tex]
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

[tex]a^m\cdot a^n = a^{m+n}\neq a^{mn}[/tex]
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:
  • #4
Mentallic
Homework Helper
3,798
94
What is

[tex]\log{a^m}+\log{a^n}[/tex]

equivalent to?
 
  • #5
31
1
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
Mentallic
Homework Helper
3,798
94
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. [itex]x^2-4x-4=0[/itex] 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
ehild
Homework Helper
15,427
1,827

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
 
  • #8
31
1
Oh I see, thank you!
 
  • #9
31
1
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
Mentallic
Homework Helper
3,798
94
The quadratic formula is the easiest method, but completing the square works too if you don't remember the formula.

[tex]x^2-4x-4=0[/tex]

Now add 4 to both sides.
[tex](x^2-4x+4)-4=4[/tex]

[tex](x-2)^2-4=4[/tex]

etc.

This is how the quadratic formula is derived. Begin with

[tex]ax^2+bx+c=0[/tex]

and then complete the square to solve for x.
 

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

  • Last Post
Replies
12
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
5
Views
4K
Replies
12
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
1K
Replies
17
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
9
Views
915
Top