Logarithms

[Acnhduy]

Solve:

log₅(x-1) + log₅(x-2) - log₅(x+6) = 0



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

Edited October 4, 2013 by Acnhduy

[studiot]

What methods, other than factorisation do you know for solving quadratic equations?

[Mozart]

There are basically three ways to solve a quadratic eguation.

1. by factorisation

2. completing the square method

3. using the almighty formula.

I see no lapses with you manipulations. you are nearly there.

[studiot]

 

There are basically three ways to solve a quadratic eguation.

 

FYI there is at least one more way.

 

http://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php

[Mozart]

 

FYI there is at least one more way.

 

http://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php

 

FYI there is at least one more way.

 

http://www.mathwarehouse.com/quadratic/roots/formula-sum-product-of-roots.php

It's true. Thank you Studiot. I forgot that.

[ComplexCalcBro]

log₅(x-1) + log₅(x-2) - log₅(x+6) = 0

 

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

 

(x-1)(x-2) = 0

 

therefore: x = 1; 2

 

Since the denominator can never be zero.

Edited October 14, 2013 by ComplexCalcBro

[overtone]

 

 

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

That is not correctly derived. Exp(0) = 1, not 0. The OP has it right, in the part where they "ignored the base".

 

 

For many people, btw, the quadratic formula is more easily remembered by sound - memorized as a one would a short poem or limerick - rather than by sight. If you are one of those people, say it as you are writing it - by hand, with a pencil - a few times, and it will stick.

[Amaton]

[math]\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)=0[/math]

 

There are probably more efficient ways, but the first thing I thought of doing was...

 

[math]5^{\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)}=5^0[/math]

 

Take 5 to the power of both sides. After some simplification...

 

[math]\frac{(x-1)(x-2)}{x+6}=1[/math]

 

From there, you can bring out the denominator to the other side and rearrange the equation into standard quadratic form.

Edited October 14, 2013 by Amaton