How can I prove that sqrt(x) > ln x when x > 0?
I can quickly see this graphically and intuitively, but how do I do a proof?
How about an argument like this:
the first derivatives of √x and lnx are
1/(2√x) and 1/x respectively.
Neither curve has any max/min points or points of inflection, so they are both increasing all the time
y = √x is clearly greater than y = lnx for 0 < x ≤ 1
(√(a decimal) is positive, ln(decimal less than 1) is negative)
since √x starts off by being above lnx and since both increase , √x stays above lnx , thus √x > lnx
you might want to include the second derivative test
the second derivatives are -1/(4x^(3/2)) and -1/x^2 respectively.
they are both slowing up, but continue to increase, but since x^2 > x^(3/2) for all x > 1 the lnx function is slowing up faster than the square roots function.
Thus the difference between the function actually gets larger and clearly √x > ln x
That works. Thanks so much!
To prove that √(x) > ln(x) when x > 0, you can use calculus and properties of logarithms. Here's a step-by-step proof:
Step 1: Define the function f(x) = √(x) - ln(x). We want to prove that f(x) > 0 for x > 0.
Step 2: Take the derivative of f(x) with respect to x:
f'(x) = (1/2) * x^(-1/2) - (1/x)
= x^(-1/2)/2 - 1/x
= (x - 2√(x))/(2x)
Step 3: Now, we need to find the critical points of f(x). Set f'(x) = 0:
(x - 2√(x))/(2x) = 0
Simplifying, we get:
x - 2√(x) = 0
Step 4: Solve for x using algebraic methods.
We can rewrite the equation as:
2√(x) = x
Squaring both sides, we get:
4x = x^2
Rearranging, we have:
x^2 - 4x = 0
Factoring, we find:
x(x - 4) = 0
So, there are two possible critical points: x = 0 and x = 4. However, x = 0 is not in the domain of the original inequality (x > 0), so we only consider x = 4 as a critical point.
Step 5: Determine the behavior of f(x) around the critical point.
When x < 4, both x and 2√(x) are positive, so f'(x) > 0.
When x > 4, both x and 2√(x) are positive, so f'(x) < 0.
Step 6: Analyze f(x) at x = 4:
Plug x = 4 into f(x):
f(4) = √(4) - ln(4)
= 2 - ln(4)
= 2 - 2ln(2)
Since ln(2) < 1, we have:
f(4) = 2 - 2ln(2) > 2 - 2 = 0
Step 7: Consider the behavior of f(x) for x approaching infinity:
lim(x -> ∞) f(x) = ∞ - ∞ = ∞ (by applying L'Hôpital's rule)
Step 8: Summarize the results:
From steps 5, 6, and 7, we can conclude that:
- f(x) > 0 for x > 4
- f(x) < 0 for x < 4
- lim(x -> ∞) f(x) = ∞
Therefore, this proves that √(x) > ln(x) when x > 0.
To prove that √x > ln(x) for x > 0, we can use basic calculus and the properties of logarithmic and square root functions. Here is a step-by-step guide on how to approach the proof:
Step 1: State the hypothesis and goal.
Hypothesis: x > 0
Goal: To prove that √x > ln(x)
Step 2: Understand the properties of the functions involved.
The square root function (√x) is always positive for positive values of x: √x > 0 for x > 0.
The natural logarithm function (ln(x)) is defined only for positive values of x.
Step 3: Define a new function for comparison.
Consider the function f(x) = √x - ln(x).
Step 4: Find the derivative of f(x).
To determine the behavior of f(x), we find its derivative f'(x).
f'(x) = (1/2√x) - (1/x) = (x - 2√x) / (2√x * x)
Simplifying further, f'(x) = (x - 2√x) / (2x√x)
Step 5: Determine when f'(x) > 0 or f'(x) < 0.
To find the critical points where f(x) may change its behavior, we need to solve for f'(x) = 0.
Setting the numerator equal to 0 yields x = 4.
Step 6: Analyze the signs of f'(x) in different intervals.
We divide the number line into three intervals: (-∞, 0), (0, 4), and (4, ∞).
For x ∈ (-∞, 0), f'(x) is undefined as log(x) is not defined for x ≤ 0.
For x ∈ (0, 4), f'(x) < 0.
For x ∈ (4, ∞), f'(x) > 0.
Step 7: Consider the behavior of f(x).
Since f'(x) > 0 on the interval (4, ∞), the function f(x) is increasing on this interval.
Step 8: Evaluate f(x) at the endpoints of the interval x ∈ (4, ∞).
Choose x = 4, calculate f(4) = √4 - ln(4) = 2 - ln(4) = 2 - 2ln(2) ≈ 0.6137.
Choose x = ∞, calculate the limit of f(x) as x approaches infinity.
lim(x→∞) f(x) = ∞ - ∞ = undefined.
Step 9: Draw the conclusion.
Since f(x) is increasing on the interval (4, ∞) and f(4) > 0, it follows that f(x) > 0 for x > 4 or x > 0.
Step 10: Relate f(x) back to √x and ln(x).
Since f(x) = √x - ln(x) and f(x) > 0 for x > 0, it implies that √x > ln(x) for x > 0.
Therefore, we have successfully proved that √x > ln(x) for x > 0.