If log3 2, log3 (2^x - 5), log3 (2^x - 7/2) are in an arithmetic progression, then the value of x is

Q: Wait, why do we reject the negative root here? I got $x=2$ and $x=3$. Why is $x=2$ wrong?

This is the classic 'JEE Trap'! In mathematics, the domain of $\log_b(y)$ is $y > 0$. When you found $x=2$, you solved the algebraic equation correctly, but you forgot to check if $x=2$ makes the arguments of the original logarithms positive. If you plug $x=2$ into $\log_3(2^x - 5)$, you get $\log_3(2^2 - 5) = \log_3(-1)$, which is undefined. Always check your domain!

Q: Why can't we just use the property $a + c = 2b$ directly without converting to logs?

You actually *are* using that property! The condition $2b = a + c$ for an A.P. is exactly what we applied. The 'log' part is just the form of the numbers. You cannot add the logs directly without using the property $\log a + \log b = \log(ab)$. You must combine them first to equate the arguments.

Q: Is it possible to have multiple solutions for $x$ in these types of problems?

Yes, absolutely! It is very common to get multiple roots from a quadratic equation. The key is to test each candidate solution against the constraints of the problem. In this case, $x=2$ failed the domain test, but $x=3$ passed. Always treat your solutions as 'candidates' until they pass the domain check.

Question

If $\log_3 2, \log_3 (2^x - 5), \log_3 (2^x - \frac{7}{2})$ are in an arithmetic progression, then the value of $x$ is equal to

Options:

(A) 4

(B) 2

(C) 3

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the edge of a mathematical landscape, looking at three mysterious blocks: $\log_3 2$, $\log_3 (2^x - 5)$, and $\log_3 (2^x - \frac{7}{2})$.

The problem states that these three terms are in an **Arithmetic Progression (A.P.)**.

In the language of algebra, if $a, b, c$ are in A.P., then they must satisfy the condition $2b = a + c$. This relationship serves as our North Star for the calculation.

### The Foundation: Applying the A.P. Condition

We start by setting up our equation using the middle term $\log_3 (2^x - 5)$.

Equating the double of the middle term to the sum of the first and third terms, we get:

$$2 \log_3 (2^x - 5) = \log_3 2 + \log_3 \left( 2^x - \frac{7}{2} \right)$$

This is the moment where the structure of the problem begins to reveal itself. We are balancing the logarithmic scale.

### The Logarithmic Toolkit

We now simplify the expression using standard logarithmic properties.

Using the power rule, $n \log a = \log a^n$, we move the coefficient of $2$ inside the logarithm on the left side. Using the product rule, $\log a + \log b = \log (ab)$, we merge the terms on the right side.

The equation simplifies to:

$$\log_3 (2^x - 5)^2 = \log_3 \left[ 2 \left( 2^x - \frac{7}{2} \right) \right]$$

This results in the cleaner form:

$$\log_3 (2^x - 5)^2 = \log_3 (2 \cdot 2^x - 7)$$

### The Algebraic Transformation

Since we have $\log_3$ on both sides, we equate the arguments:

$$(2^x - 5)^2 = 2 \cdot 2^x - 7$$

To simplify, we use the substitution $t = 2^x$. Our equation transforms into a quadratic:

$$(t - 5)^2 = 2t - 7$$

Expanding the left side gives $t^2 - 10t + 25 = 2t - 7$. Rearranging all terms to one side, we arrive at the quadratic equation:

$$t^2 - 12t + 32 = 0$$

### The Final Verdict: The Domain Check

Factorizing the quadratic yields $(t - 4)(t - 8) = 0$. This provides two potential values for $t$: $t = 4$ and $t = 8$.

Since $t = 2^x$, we find:
1. $2^x = 4 \Rightarrow x = 2$
2. $2^x = 8 \Rightarrow x = 3$

Before concluding, we must perform a **domain check** to ensure the arguments of the original logarithms remain positive.

If we test $x = 2$, the argument $2^x - 5$ becomes $2^2 - 5 = -1$. Since the logarithm of a negative number is undefined, $x = 2$ is an extraneous solution.

Testing $x = 3$ yields $2^3 - 5 = 3$ and $2^3 - 3.5 = 4.5$, both of which are positive.

The final valid solution is **$x = 3$**.

Animated Solution for Mathematics - Sequence and Series: If $lo g_{3} 2, lo g_{3} (2^{x} - 5), lo g_{3} (2^{x} - \frac{7}{2})$ are in an arithmetic progression, then the value of $x$ is equal to

Select Answer:

Visualized Solution

Identify the Arithmetic Progression

Apply the A.P. Condition

Utilize Logarithmic Properties

Simplify the Logarithmic Equation

Remove Logarithms

Substitute $t = 2^{x}$

Expand and Rearrange

Form the Quadratic Equation

Factorize and Solve for $t$

Back-Substitute to find $x$

Check the Domain (JEE Trap!)

Final Conclusion

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Foundation

The Logarithmic Toolkit

The Algebraic Transformation

The Final Verdict

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If log3​2,log3​(2x−5),log3​(2x−27​) are in an arithmetic progression, then the value of x is equal to