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Solve each inequality for x.

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

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02:24

Jeffrey Payo

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Heather Zimmers

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Functions and Models

Section 5

Inverse Functions and Logarithms

Functions

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Partial Derivatives

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Alright, it's time to solve some inequalities with exponential and log functions. This looks like fun. Alright, so let's take a look at a um we want to get the X out of the exponents so we can do that by doing the inverse function to E which is Ln Ln of every term. So we'll get L out of one Less than Ln of E to the three X -1 less. An Ln of two. Ln of one by the way is zero. Um Alright, so we get zero lesson member Ln of E. These are composite inverse functions. They undo each other mathematically and out pops the arguments. So we'll get three x minus one lesson Ln of two. Let's add one to every part. So that will give us L N two plus one. That will divide everything by three. So we get one third less than X less than Ln of two plus 1/3. So um as lovely. Let's fix that. Three. Make it look a little bit nicer. So that is then uh we were able to solve our inequality so we know what X has to be in between. So that worked well. Alright, awesome. Now we're gonna do part B, part B s, logs. So let's go ahead and start working. This first thing we'll do is subtract one from both sides. So we'll get -2. less than two when you divide by a negative. That changes the inequality. So we're going to divide by -2. That will give us Ln of X Greater than -1. Now we can do uh the inverse function that will get our argument out. So we're going to do to the L. N. Effects Greater than Edith and -1. So therefore access greater than E to the minus one and either minus one. We can write as one over E. So X is greater than one over E. So we have accomplished our task. Alright, have a great day. See you next time.

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