Set Inequality Equal Zero: First, let's set the inequality to equal zero to find the critical points. lnx−x1=0
Solve for x: Now, we need to solve for x. Let's isolate lnx.lnx=x1
Exponentiate Both Sides: To get rid of the natural log, we'll exponentiate both sides with base e.elnx=ex1
Find Approximate Solution: Since elnx is just x, we have:x=ex1
Test Critical Points: This equation isn't easy to solve algebraically, so we'll look at it graphically or use numerical methods to find the approximate value of x.
Test x<1.763: By graphing or using numerical methods, we find that the approximate solution is x≈1.763
Test x>1.763: Now we need to test intervals around our critical point to see where the inequality holds true.
Final Solution: Let's test a value less than 1.763, say x=1. We get ext{ln}(1) - rac{1}{1} = 0 - 1, which is less than 0. So the inequality doesn't hold for x<1.763.
Final Solution: Let's test a value less than 1.763, say x=1. We get ext{ln}(1) - rac{1}{1} = 0 - 1, which is less than 0. So the inequality doesn't hold for x<1.763.Now let's test a value greater than 1.763, say x=2. We get ext{ln}(2) - rac{1}{2}, which is greater than 0. So the inequality holds for x>1.763.
Final Solution: Let's test a value less than 1.763, say x=1. We get ext{ln}(1) - rac{1}{1} = 0 - 1, which is less than 0. So the inequality doesn't hold for x<1.763. Now let's test a value greater than 1.763, say x=2. We get ext{ln}(2) - rac{1}{2}, which is greater than 0. So the inequality holds for x>1.763. Therefore, the solution to the inequality x=10 is x>1.763.
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