Let S_n(x) = log_{a^(1/2)} x + log_{a^(1/3)} x + log_{a^(1/6)} x + log_{a^(1/11)} x + log_{a^(1/18)}

Q: Wait, why does the base exponent $1/k$ become a multiplier $k$?

This is a fundamental property of logarithms: $\log_{b^n} x = \frac{1}{n} \log_b x$. In our case, the base is $a^{1/k^2}$ (or similar). When you take the reciprocal of the exponent $1/k^2$, you get $k^2$. It is a powerful tool that transforms a complex-looking base into a simple coefficient, allowing us to turn a logarithmic series into a standard algebraic summation.

Q: How do I know the general term is quadratic?

Great question! Whenever the first differences of a sequence form an arithmetic progression (like $1, 3, 5, 7, \dots$), the sequence itself is quadratic. If the second differences were constant, it would be a quadratic. Here, the first differences are $1, 3, 5, 7$, which is an AP. This is a classic pattern recognition technique in JEE problems.

Q: Why can't we just use $\log_a(2x) = 2 \log_a x$?

Ah, be careful! The property $\log_a(x^n) = n \log_a x$ applies to the exponent of the *argument*, not the coefficient of the argument. For $\log_a(2x)$, we must use the product rule: $\log_a(2x) = \log_a 2 + \log_a x$. This is the key to separating the variables and solving for $a$.

Question

Let $S_n(x) = \log_{a^{1/2}} x + \log_{a^{1/3}} x + \log_{a^{1/6}} x + \log_{a^{1/11}} x + \log_{a^{1/18}} x + \log_{a^{1/27}} x + \dots$ up to $n$-terms, where $a > 1$. If $S_{24}(x) = 1093$ and $S_{12}(2x) = 265$, then value of $a$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of logarithms.

Let us begin by looking at the series:
$$S_n(x) = \log_{a^{1/2}} x + \log_{a^{1/3}} x + \log_{a^{1/6}} x + \dots$$

The bases are all powers of $a$. Recall the identity $\log_{b^k} x = \frac{1}{k} \log_b x$. By applying this, we can pull those fractional exponents out of the base and into the light as multipliers.

Suddenly, the series transforms into:
$$S_n(x) = (2 + 3 + 6 + 11 + \dots) \log_a x$$

The complexity vanishes, leaving us with a simple sum of coefficients.

### The Detective Work: Unmasking the Sequence

Now, look at those coefficients: $2, 3, 6, 11, 18, 27, \dots$. In mathematics, randomness is often just a pattern we haven't identified yet.

Let us find the differences between consecutive terms:
$3-2=1$, $6-3=3$, $11-6=5$, $18-11=7$.

The differences are $1, 3, 5, 7, \dots$, which is an arithmetic progression of odd numbers. Because the first differences form an AP, we know the general term $T_k$ must be a quadratic expression.

Testing the values:
$T_1 = 2 = 0^2 + 2$
$T_2 = 3 = 1^2 + 2$
$T_3 = 6 = 2^2 + 2$

The pattern is clear: $T_k = (k-1)^2 + 2$. We have cracked the code.

### The Summation Symphony

With the general term $T_k = (k-1)^2 + 2$, we need to sum these up to $n$ terms. Let $C_n = \sum_{k=1}^n [(k-1)^2 + 2]$.

We can split this into two manageable pieces:
$$\sum_{k=1}^n (k-1)^2 + \sum_{k=1}^n 2$$

The first part is the sum of squares from $0$ to $n-1$, which is:
$$\frac{(n-1)n(2n-1)}{6}$$

The second part is simply $2n$. Thus, our total coefficient sum is:
$$C_n = \frac{(n-1)n(2n-1)}{6} + 2n$$

### The Final Resolution

We are given two conditions. First, $S_{24}(x) = 1093$. Plugging $n=24$ into our formula:
$$C_{24} = \frac{23 \cdot 24 \cdot 47}{6} + 48 = 4324 + 48 = 4372$$

So, $4372 \log_a x = 1093$, which simplifies beautifully to $\log_a x = \frac{1}{4}$.

Second, $S_{12}(2x) = 265$. Plugging $n=12$ into our formula:
$$C_{12} = \frac{11 \cdot 12 \cdot 23}{6} + 24 = 506 + 24 = 530$$

So, $530 \log_a (2x) = 265$, which gives $\log_a (2x) = \frac{1}{2}$. Using the product rule:
$$\log_a 2 + \log_a x = \frac{1}{2}$$

Substituting $\log_a x = \frac{1}{4}$, we find $\log_a 2 = \frac{1}{4}$. Converting to exponential form, $a^{1/4} = 2$, which means **$a = 16$**.

We have arrived at the destination. The complexity was just a mask for an elegant, logical progression.

Enter Numerical Value:

Visualized Solution

Simplifying the Logarithmic Series $S_{n} (x)$

Analyzing the Coefficient Sequence

Finding the General Term $T_{k}$

Summation of the Coefficients $C_{n}$

Evaluating the Summation

Applying the First Condition: $S_{24} (x) = 1093$

Simplifying the First Condition

Applying the Second Condition: $S_{12} (2 x) = 265$

Simplifying the Second Condition

Solving for $a$ : Expanding the Logarithm

Solving for $a$ : Substitution

Final Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Detective Work

The Summation Symphony

The Final Resolution

Similar Questions

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Enter Numerical Value:

Visualized Solution

Simplifying the Logarithmic Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​(x)

Analyzing the Coefficient Sequence

Finding the General Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tk​

Summation of the Coefficients .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Cn​

Evaluating the Summation

Applying the First Condition: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S24​(x)=1093

Simplifying the First Condition

Applying the Second Condition: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S12​(2x)=265

Simplifying the Second Condition

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a: Expanding the Logarithm

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a: Substitution