Problem

(a) We must have $e^{x}-3>0 \Leftrightarrow e^{x}…

02:22

Problem 56 Hard Difficulty

Solve each inequality for x.

(a) $ 1< e^{3x - 1} < 2 $
(b) $ 1 - 2 \ln x < 3 $

Answer

a) $\frac{1}{3} < x < \frac{1+\ln 2}{3}$
b) $x > \frac{1}{c}$

Missouri State University

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Video Transcript

okay to solve these inequalities. I'm going to start with the first inequality and I'm going to take the natural log of all three parts. So we have natural log of one is less than natural law gov to the three x minus. One is less than natural log of to now in the middle natural log of Ito. A power is just going to be that power because natural log and each of the power or in vs and they cancel. So in the middle, we just have three X minus one Now on the left side. We know that natural log of one is zero, right, because each of the zero is one and then on the right, we just have natural log of to Okay, let's go ahead and add one to all three parts and we get one is less than three. X is less than natural log to. And then we're going to divide all three parts by three and we get one. Third is less than X is less than natural log of two divided by three. Now let's work on the other one. So let's start by subtracting one from both sides and we have negative to natural. Log X is less than two now. We divide both sides by negative to and remember with inequalities. When you divide by a negative, you have to reverse the signs. So now it's a greater than Okay, finally, let's exponentially eight. Which means each side becomes an exponents. Let's make each side and exponents on E so we hav e to the natural log of X is greater than e to the negative one e to the natural log of X is just X. Since inverse functions cancel so X is greater than e to the negative one. And if we want to, we can write that as one over e.

Heather Z.

Oregon State University

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Problem

(a) We must have $e^{x}-3>0 \Leftrightarrow e^{x}…

Problem 56 Hard Difficulty

Solve each inequality for x.

(a) $ 1< e^{3x - 1} < 2 $
(b) $ 1 - 2 \ln x < 3 $

Answer

a) $\frac{1}{3} < x < \frac{1+\ln 2}{3}$
b) $x > \frac{1}{c}$

More Answers

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Calculus: Early Transcendentals

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a) $\frac{1}{3} < x < \frac{1+\ln 2}{3}$
b) $x > \frac{1}{c}$