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Solve each equation for x.

(a) $ \ln (x^2 - 1) = 3 $

(b) $ e^{2x} - 3e^x + 2 = 0 $

a) $x=\sqrt{e^{3}+1}$ Or $x=-\sqrt{e^{3}+1}$

b) $x=0$ and $/$ or $x=\ln 2$

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Jp R.

September 11, 2019

Toward the end of part a you forgot to take the square root of the right side of the equation.

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

all right. We want to take our lager with Mickey Equation, and we want to change it to exponential form in order to solve it. And remember that when you see natural log, it means long based E. So we have law based E g of X squared minus one equals three. So changing that into its exponential form, we will have the base e raised to the power three equals X squared minus one. Now let's sell that for X so we can add one to both sides and we get e cubed. Plus one equals X squared. And then we can square root both sides and we get plus or minus the square root of E cubed plus one equals X and after party. I don't know about you, but when I look at this, I see a quadratic type situation. I think it looks a lot like quadratic equations that we would solve by factoring. So let's factor this into to buy no meals. To get each of the two X, you would multiply E to the X by each of the X. Remember the exponents property? You would add those powers. X plus X is two X and for the last to get to. We could do minus two and minus one and notice if you foil that, you get minus three e to the X in the inside and the outside added together. So now we'll take each of these factors and will set them equal to zero. Each of the X minus two equals zero or e to the X minus. One equals zero for the 1st 1 Add two to both sides and for the 2nd 1 add one to both sides. Okay. Now, to solve these, we're going to need to take the natural log of both sides and we get X equals Natural log of to or X equals natural log of one natural log of one is just zero. So we have X equals natural log of to or X equal zero