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Chapter 3: Problem 234

For \(x \in R, 3^{72}(1 / 3)^{x}(1 / 3)^{\sqrt{x}}>1\) then (a) \(x \in[0,64]\) (b) \(x \in(0,64)\) (c) \(x \in[0,64)\) (d) None of these

Short Answer

Expert verified
(d) None of these

Step by step solution

01

Simplify the inequality

First, let's simplify the given inequality: \(3^{72}(1 / 3)^{x}(1 / 3)^{\sqrt{x}}>1\) Since \(1/3 = 3^{-1}\), we can rewrite the inequality as: \(3^{72}3^{-x}3^{-\sqrt{x}}>1\) Now, we can combine the exponential terms: \(3^{(72-x-\sqrt{x})}>1\)
02

Eliminate the exponent

Now, let's eliminate the exponent by taking the natural logarithm (ln) of both sides: \(\ln{(3^{(72-x-\sqrt{x})})}>\ln{1}\) The logarithmic property states that \(\ln{a^b} = b\ln{a}\). Therefore, we have: \((72-x-\sqrt{x})\ln{3}>0\) Since \(\ln{3}\) is a positive number, the inequality doesn't change: \(72-x-\sqrt{x}>0\)
03

Solve for x

Now, move the x terms to the right side: \(\sqrt{x}>72-x\) Now let's square both sides to get rid of the square root: \((\sqrt{x})^2 > (72-x)^2\) \(x > 72^2 - 144x + x^2\) Next, move all the terms to the left side: \(0 > x^2 - 145x + 72^2\) Now, let's find the range of x for which the inequality holds: \(0 > (x - 64)(x - 81)\) Here, we see that the inequality holds when x is between 64 and 81. Now, let's compare this range to the given options.
04

Compare the range to the given options

Looking at the options: (a) x ∈ [0,64] (b) x ∈ (0,64) (c) x ∈ [0,64) (d) None of these None of the options provided match the range that we found (x ∈ (64,81)), so the correct answer is: (d) None of these

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Most popular questions from this chapter

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