how many solutions does an equation have?

Question

How many solutions does the following equation have?

ln(x^2 + 3x − 4) = 0

1) none

2) 1

3)2

4) infinite

I don't know how to go about solving this. I now log functions have an infinite domain so they have infinite solutions but I don't know if that applies to ln. would it be #4?

 

 

Answer 1

In order for ln(x^2 + 3x − 4) to be equal to 0, we must have x^2 + 3x − 4=e^0 and e^0=1. So you are asked to fine the number of solutions to x^2 + 3x − 4=1, or equivalently x^2 + 3x - 5=0. For this quadratic polynomial we have 3^2-4(-5)=29>0, and hence there are two solutions.

Answer 2

THE ARGUMENT IS A QUADRATIC IN X. It must be equal to 1 if ln(argument) = 0.

 

Solve x² + 3x - 4 = 1.  In other words, solve x² + 3x - 5 =0.  The Discriminant is 9 - 4(1)(-5) = 29, so there are two Real solutions.

Answer 3

Recall that if log_ax = b, then x = a^b

 

ln is an abbreviation for log base e

 

So, if ln(x²+3x-4) = 0, then x²+3x-4 = e⁰ = 1

 

Thus,  x²+3x-5 = 0

 

x = [-3±√29]/2 ≈1.2, -4.2 

 

There are 2 real (irrational) solutions.