How many solutions does the following equation have?
ln(x^{2} + 4x − 5) = 0

Question

How many solutions does the following equation have?
ln(x^{2} + 4x − 5) = 0

zepdrix · Answer

$$\large \ln\left(x^2+4x-5\right)=0$$
Hmm ok this problem is a little strange.
Here is one way to approach it at least.

Let's exponentiate both sides,
(rewrite both sides with a base of e)

$$\huge e^{\ln\left(x^2+4x-5\right)}=e^0$$
Since the exponential and the logarithm are inverse operations of one another, they essentially cancel out. Giving us,$$\large x^2+4x-5=e^0$$e^0 is just 1.$$\large x^2+4x-5=1$$Subtract 1 from each side,$$\large x^2+4x-6=0$$

If you throw this into the Quadratic Formula you'll be able to find out how many solutions exist. :D

If that's way too confusing you can let me know.
Maybe there is another way you're suppose to approach this.

anonymous · Answer

oh my gosh @zepdrix  i get it =P  that is strange