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ln x-(1)/(x) > 0

lnx1x>0 \ln x-\frac{1}{x}>0

Full solution

Q. lnx1x>0 \ln x-\frac{1}{x}>0
  1. Set Inequality Equal Zero: First, let's set the inequality to equal zero to find the critical points. lnx1x=0\ln x - \frac{1}{x} = 0
  2. Solve for x: Now, we need to solve for x. Let's isolate lnx\ln x.lnx=1x\ln x = \frac{1}{x}
  3. Exponentiate Both Sides: To get rid of the natural log, we'll exponentiate both sides with base ee.\newlineelnx=e1xe^{\ln x} = e^{\frac{1}{x}}
  4. Find Approximate Solution: Since elnxe^{\ln x} is just xx, we have:\newlinex=e1xx = e^{\frac{1}{x}}
  5. 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 xx.
  6. Test x<1.763x < 1.763: By graphing or using numerical methods, we find that the approximate solution is x1.763x \approx 1.763
  7. Test x>1.763x > 1.763: Now we need to test intervals around our critical point to see where the inequality holds true.
  8. Final Solution: Let's test a value less than 1.7631.763, say x=1x = 1. We get ext{ln}(1) - rac{1}{1} = 0 - 1, which is less than 00. So the inequality doesn't hold for x<1.763x < 1.763.
  9. Final Solution: Let's test a value less than 1.7631.763, say x=1x = 1. We get ext{ln}(1) - rac{1}{1} = 0 - 1, which is less than 00. So the inequality doesn't hold for x<1.763x < 1.763.Now let's test a value greater than 1.7631.763, say x=2x = 2. We get ext{ln}(2) - rac{1}{2}, which is greater than 00. So the inequality holds for x>1.763x > 1.763.
  10. Final Solution: Let's test a value less than 1.7631.763, say x=1x = 1. We get ext{ln}(1) - rac{1}{1} = 0 - 1, which is less than 00. So the inequality doesn't hold for x<1.763x < 1.763. Now let's test a value greater than 1.7631.763, say x=2x = 2. We get ext{ln}(2) - rac{1}{2}, which is greater than 00. So the inequality holds for x>1.763x > 1.763. Therefore, the solution to the inequality x=1x = 100 is x>1.763x > 1.763.

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