Question:
If \( \log_3 2 \), \( \log_3 (2^x - 5) \), and \( \log_3 \left( \frac{2^x - 7}{2} \right) \) are in AP, then \( x \) is equal to
If \( \log_3 2 \), \( \log_3 (2^x - 5) \), and \( \log_3 \left( \frac{2^x - 7}{2} \right) \) are in AP, then \( x \) is equal to
Show Hint
When solving logarithmic equations, use logarithmic properties to simplify and then exponentiate both sides to eliminate the logarithms.
Updated On: Apr 22, 2026
- \( 1, \frac{1}{2} \)
- \( 1, \frac{1}{3} \)
- \( 1, \frac{3}{2} \)
- None of these
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Understand the condition for terms to be in AP.
For three terms to be in arithmetic progression (AP), the middle term must be the average of the first and third terms. That is, for the terms \( a \), \( b \), and \( c \) to be in AP: \[ 2b = a + c \] In this case, the three terms are \( \log_3 2 \), \( \log_3 (2^x - 5) \), and \( \log_3 \left( \frac{2^x - 7}{2} \right) \).
Step 2: Set up the equation for AP.
Using the condition for AP, we can write: \[ 2 \log_3 (2^x - 5) = \log_3 2 + \log_3 \left( \frac{2^x - 7}{2} \right) \]
Step 3: Apply logarithmic properties.
We can use the property of logarithms that \( \log_b a + \log_b c = \log_b (ac) \). So, the right-hand side becomes: \[ \log_3 2 + \log_3 \left( \frac{2^x - 7}{2} \right) = \log_3 \left( 2 \times \frac{2^x - 7}{2} \right) = \log_3 (2^x - 7) \] Thus, the equation becomes: \[ 2 \log_3 (2^x - 5) = \log_3 (2^x - 7) \]
Step 4: Eliminate the logarithms.
We can eliminate the logarithms by exponentiating both sides. Since the logarithms have the same base (3), we can drop the logs and equate the arguments: \[ (2^x - 5)^2 = 2^x - 7 \]
Step 5: Expand and simplify.
Expanding the left-hand side: \[ (2^x - 5)^2 = 2^{2x} - 10 \cdot 2^x + 25 \] Thus, the equation becomes: \[ 2^{2x} - 10 \cdot 2^x + 25 = 2^x - 7 \] Simplifying further: \[ 2^{2x} - 10 \cdot 2^x + 25 = 2^x - 7 \] Now move all terms to one side: \[ 2^{2x} - 11 \cdot 2^x + 32 = 0 \]
Step 6: Solve the quadratic equation.
Let \( y = 2^x \). Substituting this into the equation: \[ y^2 - 11y + 32 = 0 \] Solve this quadratic equation using the quadratic formula: \[ y = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(32)}}{2(1)} = \frac{11 \pm \sqrt{121 - 128}}{2} = \frac{11 \pm \sqrt{-7}}{2} \] Since the discriminant is negative, there are no real solutions for \( y \), which implies that the equation does not have a real solution for \( x \).
Step 7: Conclusion.
Thus, there is no real value for \( x \) that satisfies the given condition, corresponding to option (D), "None of these."
For three terms to be in arithmetic progression (AP), the middle term must be the average of the first and third terms. That is, for the terms \( a \), \( b \), and \( c \) to be in AP: \[ 2b = a + c \] In this case, the three terms are \( \log_3 2 \), \( \log_3 (2^x - 5) \), and \( \log_3 \left( \frac{2^x - 7}{2} \right) \).
Step 2: Set up the equation for AP.
Using the condition for AP, we can write: \[ 2 \log_3 (2^x - 5) = \log_3 2 + \log_3 \left( \frac{2^x - 7}{2} \right) \]
Step 3: Apply logarithmic properties.
We can use the property of logarithms that \( \log_b a + \log_b c = \log_b (ac) \). So, the right-hand side becomes: \[ \log_3 2 + \log_3 \left( \frac{2^x - 7}{2} \right) = \log_3 \left( 2 \times \frac{2^x - 7}{2} \right) = \log_3 (2^x - 7) \] Thus, the equation becomes: \[ 2 \log_3 (2^x - 5) = \log_3 (2^x - 7) \]
Step 4: Eliminate the logarithms.
We can eliminate the logarithms by exponentiating both sides. Since the logarithms have the same base (3), we can drop the logs and equate the arguments: \[ (2^x - 5)^2 = 2^x - 7 \]
Step 5: Expand and simplify.
Expanding the left-hand side: \[ (2^x - 5)^2 = 2^{2x} - 10 \cdot 2^x + 25 \] Thus, the equation becomes: \[ 2^{2x} - 10 \cdot 2^x + 25 = 2^x - 7 \] Simplifying further: \[ 2^{2x} - 10 \cdot 2^x + 25 = 2^x - 7 \] Now move all terms to one side: \[ 2^{2x} - 11 \cdot 2^x + 32 = 0 \]
Step 6: Solve the quadratic equation.
Let \( y = 2^x \). Substituting this into the equation: \[ y^2 - 11y + 32 = 0 \] Solve this quadratic equation using the quadratic formula: \[ y = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(32)}}{2(1)} = \frac{11 \pm \sqrt{121 - 128}}{2} = \frac{11 \pm \sqrt{-7}}{2} \] Since the discriminant is negative, there are no real solutions for \( y \), which implies that the equation does not have a real solution for \( x \).
Step 7: Conclusion.
Thus, there is no real value for \( x \) that satisfies the given condition, corresponding to option (D), "None of these."
Was this answer helpful?
0
0
Top MET Mathematics Questions
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- The equation $3^{3x + 4} = 9^{2x - 2},\ x>0$ has the solution
- A parallelogram is constructed on the vectors $\mathbf{a} = 3\alpha - \beta$, $\mathbf{b} = \alpha + 3\beta$. If $|\alpha| = |\beta| = 2$ and the angle between $\alpha$ and $\beta$ is $\dfrac{\pi}{3}$, then the length of a diagonal of the parallelogram is
- Every group of order 7 is
- Let $U_{n} = 2 + 2^{3} + 2^{5} + \cdots + 2^{2n+1}$ and $V_{n} = 1 + 4 + 4^{2} + \cdots + 4^{n-1}$. Then $\displaystyle\lim_{n \to \infty} \dfrac{U_n}{V_n}$ is equal to
View More Questions
Top MET sequences Questions
- Let $U_{n} = 2 + 2^{3} + 2^{5} + \cdots + 2^{2n+1}$ and $V_{n} = 1 + 4 + 4^{2} + \cdots + 4^{n-1}$. Then $\displaystyle\lim_{n \to \infty} \dfrac{U_n}{V_n}$ is equal to
- The sum to $n$ terms of the series $\dfrac{1}{2} + \dfrac{3}{4} + \dfrac{7}{8} + \dfrac{15}{16} + \cdots$ is}
- The value of $\displaystyle\sum_{r=1}^{\infty}\left[3\cdot 2^{-r} - 2\cdot 3^{1-r}\right]$ is
- The sum of infinite terms of the GP $\dfrac{\sqrt{2}+1}{\sqrt{2}-1},\;\dfrac{1}{2-\sqrt{2}},\;\dfrac{1}{2},\;\ldots$ is
- $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)}$ is equal to
View More Questions
Top MET Questions
- A coil of 100 turns and area $2 \times 10^{-2}m^2 $ is pivoted about a vertical diameter in a uniform magnetic field and carries a current of 5A. When the coil is held with its plane in north-south direction, it experiences a couple of 0.33 Nm. When the plane is east-west, the corresponding couple is 0.4 Nm, the value of magnetic induction is [Neglect earth's magnetic field]
- $\int\limits_{0}^{8}| x -5| d x$ is equal to
- Radiation, with wavelength 6561 $?$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} T$. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
- In intrinsic semiconductor at room temperature number of electrons and holes are
- Which of the following product is formed in the reaction $ CH_3MgBr {->[(i)CO_2][(ii)H_2O]} ? $
View More Questions