i dont get this log. ...?

Question

log[1/3](x^2+x) - log[1/3](x^2-x) = -1


.. i asked this question before but im still confused! if you could, please write out what you are doing in each step! thanksss so much :)

btw, the answer is 2.

btw, the answer is 2.

Answer 1

log[1/3](x^2+x) - log[1/3](x^2-x) = -1

I'm assuming you are using [ ] to represent the base of the log and that you are solving for x.

log[1/3](x^2+x) - log[1/3](x^2-x) = -1
Combing the logs:
log[1/3]((x^2+x)/(x^2-x)) = -1

Important reference:
log [a](b) = c
a^c = b


Using this we can rewrite our problem:

log[1/3]((x^2+x)/(x^2-x)) = -1
(1/3)^(-1) = (x^2+x)/(x^2-x)
3 = (x^2+x)/(x^2-x)
3*(x^2-x) = (x^2+x)
3x^2-3x = x^2+x
2x^2-4x = 0
2x(x-2) = 0
x = 0, 2

Test each answer, since we cannot have the log of 0, x=0 is an extraneous answer.

The answer is x = 2.

Answer 2

We start with this:

1/3 = 3^-1

Now we have:

Log.1/3(x^2 + x) = Log.3^-1 ( x^2 + x) = -Log.3(x^2 + x)


The next logarithm will be:

-Log.1/3 ( x^2 - x) = -Log.3^-1 ( x^2 - x) = +Log.3 ( x^2 - x)

So we have

+Log.3(x^2 - x) - Log.3(x^2 + x) = -1

The bases of the logarithms are the same so we can divide them (because of the - sign)

(x^2 - x) / (x^2 + x) = -1

Multiply the denominator by -1:

x^2 - x = -x^2 - x

This means that x can be any number given the fact that a negative number on power of an even number results in a positive number.

Good luck

Answer 3

I suppose that the base is 1/3
so log(x^2+x)-log( x^2-x) is log[(x^2+x)/((x^2-x)]=-1
so the number is the base raised to (-1)
(x^2+x)/(x^2-x) =1/3^-1 =3
x^2+x=3x^2-3x 2x^2-4x= 0
x(2x-4)=0 x=0(no solution zero has NO log
2x-4=0 x=2 which is solution

Answer 4

log(base1/3) (x^2 + x) – log(base1/3) (x^2 – x) = -1 (combine the logs since they have the same base)
log(base 1/3) [(x^2 + x)/(x^2 – x)] = -1

…and that’s all I know how to do up to…