The value of 6 + log₃/₂ ( (1/(3√2)) √(4 - (1/(3√2)) √(4 - (1/(3√2)) √(4 - ...))) ) is

Q: Wait, why can we just replace the infinite part with $y$? Doesn't that change the value?

That is a brilliant question! Think of it this way: if you have an infinite ocean and you take away one drop, you still have an infinite ocean. Because the expression is infinite, the nested part inside the first square root is identical to the whole expression. By setting it equal to $y$, we are capturing that self-similarity mathematically. It is not changing the value; it is defining the structure of the infinity.

Q: Why do we reject the negative root when solving the quadratic equation?

Remember the definition of the principal square root function, $\sqrt{x}$. By convention, the symbol $\sqrt{\cdot}$ denotes the non-negative root. Since our variable $y$ is defined as a square root, it must be greater than or equal to zero. Therefore, any negative solution we get from the quadratic formula is physically impossible in this context and must be discarded.

Q: How do I handle the base change in the logarithm? The base is $3/2$ but the argument is $4/9$.

This is a classic JEE manipulation! Notice that $4/9$ is the square of $2/3$. Since our base is $3/2$, we can write $4/9$ as $(2/3)^2$. To make the base match, we use the property that $(2/3) = (3/2)^{-1}$. Thus, $(2/3)^2 = (3/2)^{-2}$. Now, the logarithm becomes $\log_{3/2} ((3/2)^{-2})$, which simplifies beautifully to $-2$.

Question

The value of $6 + \log_{3/2} \left( \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \dots}}} \right)$ is

Sigma Percentile · Accepted Answer

### The Infinite Puzzle: A Journey into Nested Radicals

Have you ever looked at a problem and felt like it was staring back at you, daring you to blink? This problem is exactly that—a beautiful, intimidating, infinite nested radical wrapped in a logarithm.

It looks like a labyrinth, but I promise you, there is a clear path through it. Let's walk this path together.

### Phase 1: The Infinite Sea

Look closely at the expression:

$$6 + \log_{\frac{3}{2}} \left( \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \dots}}} \right)$$

It is easy to get lost in the dots, but the secret to solving infinite structures is to find the **self-similarity**. Imagine you are standing on the edge of this infinite sequence; if you were to peel away the very first layer, you would see the exact same infinite pattern staring back at you.

Let's define this entire repeating nested radical as a single variable, $y$:

$$y = \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \frac{1}{3\sqrt{2}} \sqrt{4 - \dots}}}$$

Because the pattern is infinite, we can replace the inner repeating part with $y$ itself. This transforms our terrifying infinite expression into a simple, elegant recursive equation:

$$y = \sqrt{4 - \frac{1}{3\sqrt{2}} y}$$

### Phase 2: The Quadratic Transformation

Now that we have our equation, we need to strip away that square root. Squaring both sides gives:

$$y^2 = 4 - \frac{y}{3\sqrt{2}}$$

To make this look like a standard quadratic equation, let's clear the denominator by multiplying the entire equation by $3\sqrt{2}$:

$$3\sqrt{2}y^2 = 12\sqrt{2} - y$$

Rearranging everything to one side, we arrive at:

$$3\sqrt{2}y^2 + y - 12\sqrt{2} = 0$$

This is a standard quadratic equation of the form $ay^2 + by + c = 0$, where $a = 3\sqrt{2}$, $b = 1$, and $c = -12\sqrt{2}$.

### Phase 3: The Discriminant & The Choice

Let's calculate the discriminant, $D = b^2 - 4ac$. Substituting our values:

$$D = (1)^2 - 4(3\sqrt{2})(-12\sqrt{2})$$

Calculating this carefully:

$$D = 1 - 4 \cdot (-36) \cdot 2 = 1 + 288 = 289$$

Since $289$ is a perfect square ($17^2$), we use the quadratic formula $y = \frac{-b \pm \sqrt{D}}{2a}$ to find:

$$y = \frac{-1 \pm 17}{6\sqrt{2}}$$

This gives us two potential values for $y$: $y = \frac{16}{6\sqrt{2}}$ or $y = \frac{-18}{6\sqrt{2}}$.

Because $y$ is defined as a principal square root, it must be non-negative. We reject the negative root and accept:

$$y = \frac{16}{6\sqrt{2}} = \frac{8}{3\sqrt{2}}$$

### Phase 4: The Logarithmic Finale

We are almost there. Let's substitute $y$ back into our original expression:

$$6 + \log_{\frac{3}{2}} \left( \frac{1}{3\sqrt{2}} \cdot \frac{8}{3\sqrt{2}} \right)$$

Simplifying the argument, the numerator is $8$ and the denominator is $(3\sqrt{2}) \cdot (3\sqrt{2}) = 9 \cdot 2 = 18$. Thus, the argument is $\frac{8}{18} = \frac{4}{9}$.

Now we evaluate the logarithm:

$$\log_{\frac{3}{2}} \left( \frac{4}{9} \right) = \log_{\frac{3}{2}} \left( \left( \frac{2}{3} \right)^2 \right) = \log_{\frac{3}{2}} \left( \left( \frac{3}{2} \right)^{-2} \right) = -2$$

Finally, we add the $6$ from the beginning:

$$6 + (-2) = 4$$

The complexity collapses, the infinite sequence resolves, and we are left with a clean, satisfying integer. The final answer is **4**.

Animated Solution for Mathematics - Quadratic Equations: The value of $6 + lo g_{3/2} (\frac{1}{3 2} 4 - \frac{1}{3 2} 4 - \frac{1}{3 2} 4 - \dots)$ is

Enter Numerical Value:

Visualized Solution

Identify the Nested Structure

Define a Variable for the Infinite Part

Form the Recursive Equation

Remove the Square Root

Convert to Standard Quadratic Form

Identify Coefficients and Discriminant

Calculate the Discriminant

Find the Roots

Select the Valid Root

Substitute Back into the Main Expression

Simplify the Argument of the Logarithm

Express as a Power of the Base

Final Calculation

The Sigma Insight: Solution of Quadratic Equations

The Infinite Puzzle

Phase 1

Phase 2

Phase 3

Phase 4

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