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Chapter 8: Problem 650

If \(\log _{3} 2, \log _{3}\left(2^{x}-5\right)\) and \(\log _{3}\left[2^{x}-(7 / 2)\right]\) are in A. P. then \(\mathrm{x}=\) (A) 2 (B) 3 (C) 4 (D) 2 or 3

Short Answer

Expert verified
The value of \(x\) is either \(2\) or \(3\). So, the correct answer is (D) 2 or 3.

Step by step solution

01

Write the Arithmetic Progression Relation

Since \(\log_{3}(2)\), \(\log_{3}(2^x -5)\), and \(\log_{3}(2^x -\frac{7}{2})\) are in A. P., we can write: \(2 \cdot \log_{3}(2^x - 5) = \log_{3}(2) + \log_{3}(2^x - \frac{7}{2})\)
02

Convert Logarithmic Equation to Exponential Form

To solve for \(x\), we first need to remove the logarithmic functions from the equation. We can do this by applying the following property of logarithms: \(\log_{a}(b) = c \Leftrightarrow a^c = b\). Our equation now becomes: \(3^{2 \cdot \log_{3}(2^x - 5)} = 3^{\log_{3}(2) + \log_{3}(2^x - \frac{7}{2})}\)
03

Simplify The Equation

Now, we can use the properties of exponents to simplify the equation further: \((3^{\log_{3}(2^x - 5)})^2 = 3^{\log_{3}(2) \cdot 3^{\log_{3}(2^x - \frac{7}{2})}}\)
04

Convert Back to the Original Expression

Next, apply the property of logarithms we used earlier in reverse, so the equation becomes: \((2^x - 5)^2 = 2 \cdot (2^x - \frac{7}{2})\)
05

Expand and Simplify

Expanding the left side of the equation, we get: \(4^x - 20 \cdot 2^{x-1} + 25 = 4 \cdot 2^{x-1} - 7\) Now, let \(y = 2^{x-1} \Rightarrow 2^x = 2y\). So, we have: \(4y^2 - 20y + 25 = 4y - 7\)
06

Solve for y

Rearranging, we have a quadratic equation: \(4y^2 - 24y + 32 = 0\) Divide both sides by \(4\): \(y^2 - 6y + 8 = 0\) Factorize: \((y-2)(y-4) = 0\) Thus the possible values of \(y\) are \(2\) and \(4\).
07

Solve for x

Recall that we defined \(y = 2^{x-1}\). Hence, for respective values of \(y\), we have: 1. \(2^{x-1} = 2 \Rightarrow x - 1 = 1 \Rightarrow x = 2\) 2. \(2^{x-1} = 4 \Rightarrow x - 1 = 2 \Rightarrow x = 3\)
08

Check Options

Hence, \(x\) can be \(2\) or \(3\), which corresponds to option (D).

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Most popular questions from this chapter

If \(\mathrm{S}_{\mathrm{n}}=\mathrm{an}+\mathrm{bn}^{2}\), for an A. P. where a and \(\mathrm{b}\) are constants, then common difference of A. P. will be (A) \(2 \mathrm{~b}\) (B) \(a+b\) (C) \(2 \mathrm{a}\) (D) \(a-b\)

\(\tan ^{-1}(1 / 3)+\tan ^{-1}(1 / 7)+\tan ^{-1}(1 / 13)+\ldots+\tan ^{-1}[1 /(9703)]\) (A) \(\begin{array}{lll}\text { (B) }(\pi / 6) & \text { (C) }(\pi / 3) & \text { (D) } \tan ^{-1}(0.98)\end{array}\)

A. \(\mathrm{M}\) of the three numbers which are in G. P. is \([(14) / 3]\) If adding 1 in first and second number and subtracting 1 from the third number, resulting numbers are in \(\mathrm{A} . \mathrm{P}\). then the sum of the squares of original three numbers is (A) 91 (B) 80 (C) 84 (D) 88

\(2+12+36+80+\ldots \mathrm{n}\) terms \(=\) (A) \([\\{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)(3 \mathrm{n}+5) \overline{\\} /(24})]\) (B) \([\\{n(n+1)(n+2)(3 n+1)\\} /(12)]\) (C) \([\\{n(n+1)(n+3)(n+5)\\} /(24)]\) (D) \([\\{n(n+1)(n+2)(n+3)\\} /(12)]\)

If the pth term of a G. P. is \(x\) and \(q\) th term is \(y\), then the nth term is (A) \(\left[x^{\overline{n-p} / y^{n}-q}\right]^{[1 /(p-q)]}\) (B) \(\left[x^{n+q} / y^{n+p}\right]^{[1 /(p-q)]}\) (C) \(\left[x^{n-q} / y^{n-p}\right]^{[1 /(p-q)]}\) (D) \(\left[x^{n-q} / y^{n-p}\right]^{[1 /(p+q)]}\)
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