logarithm and domain range finding question

Q: Find the domain of each function.

$\displaystyle g(x)=\log_{5}(\displaystyle{\frac{x+1}{x}})$

now the domain here is $\displaystyle x>0 $ and $\displaystyle x<-1$

I want to know how to do calculate that?

I was thinking / trying to take x to the other side (dividing both sides with x) in this inequality $\displaystyle (\displaystyle{\frac{x+1}{x}})>0$

but that removes one x and I can't calculate the domain range as I want it. what I mean is that if I do this $\displaystyle (x+1)>0$

then $\displaystyle x>-1$ which would mean the domain is just greater than -1 which is wrong.

I hope I explained my question here.

Thanks in advance.

Re: logarithm and domain range finding question

$\displaystyle \frac{x+1}{x}>0$ iff either x + 1 > 0 and x > 0, or x + 1 < 0 and x < 0. In the first case, x + 1 > 0 gives x > -1. However, we need *both* conditions to be true: x + 1 > 0 and x > 0. The second one is more restrictive (it implies the first one). Therefore, x + 1 > 0 and x > 0 is equivalent to x > 0. Now consider the second case similarly.

Edit: Dividing both sides of an inequality by x has to be done carefully. If you divide both sides of an equation, you just need to require that x ≠ 0. If you divide both sides of an inequality, you need to consider the cases x > 0 and x < 0 since dividing by a negative number flips the inequality. Thus, one inequality is split into two systems of two inequalities. In this particular problem, it is easier to use the fact that a / b > 0 iff a and b are either both positive or both negative.

Re: logarithm and domain range finding question

Quote:

Originally Posted by

**ameerulislam** Q: Find the domain of each function.

$\displaystyle g(x)=\log_{5}(\displaystyle{\frac{x+1}{x}})$

now the domain here is $\displaystyle x>0 $ and $\displaystyle x<-1$

I want to know how to do calculate that?

I was thinking / trying to take x to the other side (dividing both sides with x) in this inequality $\displaystyle (\displaystyle{\frac{x+1}{x}})>0$

You know that $\displaystyle {\dfrac{x+1}{x}>0$ so the numerator and the denominator must have the same sign.

That happens if $\displaystyle x<-1\text{ or }x>0.$ Don't over think it.

Re: logarithm and domain range finding question

Quote:

Originally Posted by

**emakarov** $\displaystyle \frac{x+1}{x}>0$ iff either x + 1 > 0 and x > 0, or x + 1 < 0 and x < 0. In the first case, x + 1 > 0 gives x > -1. However, we need *both* conditions to be true: x + 1 > 0 and x > 0. The second one is more restrictive (it implies the first one). Therefore, x + 1 > 0 and x > 0 is equivalent to x > 0. Now consider the second case similarly.

Edit: Dividing both sides of an inequality by x has to be done carefully. If you divide both sides of an equation, you just need to require that x ≠ 0. If you divide both sides of an inequality, you need to consider the cases x > 0 and x < 0 since dividing by a negative number flips the inequality. Thus, one inequality is split into two systems of two inequalities. In this particular problem, it is easier to use the fact that a / b > 0 iff a and b are either both positive or both negative.

x>0 is one range (fine with this one) another range is x<-1, considering both numerator and denominator . coz if x is -2 (numerator) than and it divides by -2 (denominator), the result is positive which satisfies logarithms.

Re: logarithm and domain range finding question

so how to calculate that using considering both conditions.

Re: logarithm and domain range finding question

Quote:

Originally Posted by

**ameerulislam** so how to calculate that using considering both conditions.

Quote:

Originally Posted by

**emakarov** Now consider the second case similarly.

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