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Numerade Educator



Problem 55 Hard Difficulty

Solve each inequality for x.

(a) $ \ln x < 0 $
(b) $ e^x > 5 $


(a) $\ln x<0 \Rightarrow x<e^{0} \Rightarrow x<1$. Since the domain of $f(x)=\ln x$ is $x>0,$ the solution of the original inequality
is $0<x<1$.
(b) $e^{a}>5 \Rightarrow \ln e^{x}>\ln 5 \Rightarrow x>\ln 5$.

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Video Transcript

I'm going to solve both of these inequalities algebraic Lee and then confirmed them graphically. So first of all, with natural log of X is less than zero. I'm going to use a process called exponentially ating, which means that each side of the equation or each side of the inequality will be an exponents on the same base. And I'm going to use based E. So we hav e raised to the left side of the inequality is less than e raised to the right side of the inequality. And we should know, based on inverse functions that e to the natural log of X simplifies to just be X, and each of zero is one, so we have X is less than one. There's one other thing we need to consider, though. Remember the domain of along with Mick Function X has to be greater than zero. So we do have X is less than one, and we have X is greater than zero. When you combine those together, you get zero is less than X is less than one now. Graphically, if you think about what the natural log of X function looks like, it looks like this with the next intercept at one. Where is it greater than zero Down here. We're sorry. Less than zero down here that that's where the Y values are. Less than zero. And that was between X equals zero and X equals one. Okay, now let's do something similar to the other inequality. And I'm going to take the natural log of both sides. So we have natural log of e to the X is greater than natural law. Go five because of inverse functions. Natural log of each of the X simplifies to just be x so x is greater than natural log five.