# Logarithms

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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.

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

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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.

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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 by ComplexCalcBro
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[(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.

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$\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)=0$

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

$5^{\log_5 (x-1)+\log_5 (x-2)-\log_5 (x+6)}=5^0$

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

$\frac{(x-1)(x-2)}{x+6}=1$

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

Edited by Amaton

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