If log_3 2, log_3 (2^x - 5), and log_3 (2^x - 7/2) are in arithmetic progression, determine the valu

Q: Wait, I got two values for $x$, $2$ and $3$. Why do we reject $x=2$? It satisfies the quadratic equation perfectly!

That is a classic JEE trap! While $x=2$ solves the algebraic equation derived from the logs, it fails the domain requirement of the original logarithmic function. Remember, $\log_b(A)$ is only defined if $A > 0$. Plugging $x=2$ into $\log_3(2^x - 5)$ gives $\log_3(4-5) = \log_3(-1)$, which is undefined in the real number system. Always check your domain!

Q: Why did we have to use the power rule and product rule first? Can't we just equate the arguments of the three logs directly since they are in A.P.?

Great question! You cannot equate the arguments directly because the terms are in A.P., not equal to each other. The condition for A.P. is $2b = a + c$. You must apply the log properties to combine the terms *before* you can equate the arguments. Without the log rules, you are just comparing the values, not the structure of the equation.

Q: Is the substitution $y = 2^x$ really necessary? Can't I just solve it with $2^x$?

You absolutely can solve it without substitution, but it is much riskier. Working with $2^x$ directly often leads to algebraic errors, especially when squaring terms. Substitution acts as a safety net, simplifying the expression into a standard quadratic $y^2 - 12y + 32 = 0$, which is much easier to factorize and solve without losing track of the exponents.

Question

If $\log_3 2, \log_3 (2^x - 5)$, and $\log_3 (2^x - \frac{7}{2})$ are in arithmetic progression, determine the value of $x$.

Sigma Percentile · Accepted Answer

### The Logarithmic Dance: A Journey into Arithmetic Progressions

Welcome, my dear student. Today, we are going to dissect a problem that is a quintessential JEE Advanced favorite. It combines the elegance of Arithmetic Progressions (A.P.) with the strict, unforgiving nature of logarithmic domains.

Many students rush into the algebra and lose marks on the final step. We will not be those students. Let us walk through this together.

### Phase 1: The Foundation of A.P.

We are given three terms: $\log_3 2$, $\log_3 (2^x - 5)$, and $\log_3 (2^x - \frac{7}{2})$. We are told they are in an Arithmetic Progression.

This means the difference between consecutive terms is constant. If $a, b, c$ are in A.P., then $b - a = c - b$, which rearranges into the condition $2b = a + c$.

This is our golden key. Let us set up the equation:

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

### Phase 2: The Logarithmic Transformation

Now, we have an equation, but it is cluttered with logarithms. We need to simplify.

Look at the left side: $2 \log_3 (2^x - 5)$. We use the Power Rule of logarithms, which states that $n \log_b(m) = \log_b(m^n)$. That coefficient of $2$ jumps inside to become an exponent: $\log_3 (2^x - 5)^2$.

Now look at the right side: $\log_3 2 + \log_3 (2^x - \frac{7}{2})$. We use the Product Rule, $\log_b(m) + \log_b(n) = \log_b(mn)$. This condenses the right side into $\log_3 [2(2^x - \frac{7}{2})]$.

Now, our equation is perfectly balanced:

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

### Phase 3: The Algebraic Leap

Since the logarithmic function is one-to-one, if $\log_3(A) = \log_3(B)$, then $A$ must equal $B$. We can drop the logs!

This leaves us with $(2^x - 5)^2 = 2(2^x - \frac{7}{2})$. Let us use a substitution to clear the fog. Let $y = 2^x$.

Our equation transforms into $(y - 5)^2 = 2(y - \frac{7}{2})$. Expanding both sides, we get $y^2 - 10y + 25 = 2y - 7$.

Bringing everything to one side, we arrive at the quadratic equation:

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

### Phase 4: The Final Verdict

Factorizing $y^2 - 12y + 32 = 0$ is straightforward. We need two numbers that multiply to $32$ and add to $-12$. Those are $-4$ and $-8$.

So, $(y - 4)(y - 8) = 0$, giving us $y = 4$ or $y = 8$. Substituting back $y = 2^x$, we get:

$$2^x = 4 \Rightarrow x = 2$$
$$2^x = 8 \Rightarrow x = 3$$

But wait! We must perform the domain check. If $x = 2$, the term $\log_3 (2^x - 5)$ becomes $\log_3 (4 - 5) = \log_3 (-1)$, which is undefined. We must reject $x = 2$.

If $x = 3$, the term becomes $\log_3 (8 - 5) = \log_3 (3)$, which is perfectly valid. Thus, the only solution is **$x = 3$**.

Always remember: in the world of JEE, the algebra is only half the battle; the domain is where the war is won.

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

Enter Numerical Value:

Visualized Solution

Condition for Arithmetic Progression

Substituting the Logarithmic Terms

Applying the Logarithmic Power Rule

Applying the Logarithmic Product Rule

Equating the Logarithmic Arguments

Simplifying with Substitution

Expanding Both Sides

Forming the Quadratic Equation

Factorizing the Quadratic Equation

Back-Substitution for x

Checking Domain Constraints (The Trap)

Final Conclusion

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

The Logarithmic Dance

Phase 1

Phase 2

Phase 3

Phase 4

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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), and log3​(2x−27​) are in arithmetic progression, determine the value of x.