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Chapter 5: Problem 84

The minimum value of \(2 \log _{10} x-\log _{x} .01, x>1\) is (a) 1 (b) \(-1\) (c) 2 (d) None of these

Short Answer

Expert verified
The minimum value of the function \(2 \log _{10} x-\log _{x} 0.01\) for \(x > 1\) is 0, which is not present in the given options. Therefore, the correct answer is (d) None of these.

Step by step solution

01

Simplify the function using properties of logarithms

The given function is \(2 \log _{10} x-\log _{x} 0.01\). We can first simplify the function using the properties of logarithms: \(\log _{x} 0.01 = \log _{x} 10^{-2}\) Applying the change of base formula, we get: \(\frac{\log _{10} 10^{-2}}{\log _{10} x}\) Simplify again, and the function becomes: \(f(x) = 2 \log _{10} x - \frac{-2}{\log _{10} x}\)
02

Find the first derivative of the function

To find the critical points of the function and examine if it has minima, we need to first find its first derivative. \(f'(x) = \frac{d}{dx} \left( 2 \log _{10} x - \frac{-2}{\log _{10} x} \right)\) Using the chain rule, \(f'(x) = \frac{2}{x \ln{10}} - \frac{2}{(\log _{10} x)^2 \ln{10}} \cdot \frac{1}{x \ln{10}}\)
03

Find the critical points by equating the first derivative to zero

To find the critical points, we equate f'(x) to 0: \(\frac{2}{x \ln{10}} - \frac{2}{(\log _{10} x)^2 \ln{10}} \cdot \frac{1}{x \ln{10}} = 0\) Multiplying both sides by \((x \ln{10})^2\), we get: \(2(x \ln{10}) - 2 \log _{10} x = 0\) By solving for x, we get: \(x \ln{10} - \log _{10} x = 0\) Dividing by \(\ln{10}\): \(x - \frac{\log _{10} x}{\ln{10}} = 0\)
04

Find the second derivative of the function

To determine if the critical point is a minimum, find the second derivative of the function: \(f''(x) = \frac{d^2}{dx^2} \left( \frac{2}{x \ln{10}} - \frac{2}{(\log _{10} x)^2 \ln{10}} \cdot \frac{1}{x \ln{10}} \right)\) Using the chain rule again, we find that: \(f''(x) = -\frac{2}{(x \ln{10})^2} + \frac{6}{(x \ln{10})^3}\)
05

Determine if the critical point is a minimum

Find the value of the second derivative at the critical point found in Step 3: \(f''(x) > 0\), then the function has a minimum at this critical point. By numerical methods, we find out that the critical point x is approximately equal to 10. Substitute x = 10 to the second derivative: \(f''(10) > 0\) Therefore, the function has a minimum value at x = 10.
06

Calculate the minimum value

To find the minimum value of the function, substitute x = 10 into the original function: \(f(10) = 2 \log _{10} 10 - \frac{-2}{\log _{10} 10} = 2 - 2 = 0\) So, the minimum value of the function is 0, which is not present in the given options. Hence, the correct answer is: (d) None of these

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Most popular questions from this chapter

Consider \(f(x)=\int_{0}^{x}\left(t+\frac{1}{t}\right) d t\) and \(g(x)=f^{\prime}(x)\) for \(x\) \(\in\left[\frac{1}{2}, 3\right] .\) If \(P\) is a point on the curve \(y=g(x)\) such that the tangent to this curve at \(P\) is parallel to a chord joining the points \(\left(\frac{1}{2}, g\left(\frac{1}{2}\right)\right)\) and \((3, g(3))\) of the curve, then the coordinates of the point \(P\) (a) can't be found out (b) \(\left(\frac{7}{4}, \frac{65}{28}\right)\) (c) \((1,2)\) (d) \(\left(\sqrt{\frac{3}{2}}, \frac{5}{\sqrt{6}}\right)\)

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