The least value of the expression 2 log₁₀x - logₓ(0.01), for x 1, is (A) 10 (B) 2 (C) -0.01 (D) no

Q: Wait, why do we reject the negative root in the AM-GM inequality?

That is a fantastic observation! The AM-GM inequality, which states that $\frac{a+b}{2} \ge \sqrt{ab}$, is strictly valid for positive real numbers. In our case, because $x > 1$, we know that $\log_{10}x > 0$. Since our variable $t = \log_{10}x$ is strictly positive, we are safely within the domain where AM-GM applies. We don't 'reject' a negative root; rather, the domain of our function excludes negative values entirely!

Q: Can I use calculus to solve this instead of AM-GM?

Absolutely! You can define $f(t) = 2(t + \frac{1}{t})$, find the derivative $f'(t) = 2(1 - \frac{1}{t^2})$, and set it to zero to find the critical point at $t=1$. However, in the JEE Advanced, time is your most precious resource. AM-GM provides the answer in seconds, whereas calculus requires more algebraic manipulation. Both are correct, but AM-GM is the 'elegant' path.

Q: Why is the answer 'none of these'? It feels like I missed something.

This is a classic JEE psychological trap! Students often panic when their calculated answer isn't among the options. But remember, the exam tests your confidence in your derivation. If your math is sound—and here, the transformation to $2(t + \frac{1}{t})$ and the application of AM-GM are rock solid—then 'none of these' is a perfectly valid and correct answer. Trust your process!

Question

The least value of the expression $2 \log_{10}x - \log_x(0.01)$, for $x > 1$, is

Options:

(A) 10

(B) 2

(C) $-0.01$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with finding the least value of the expression $2 \log_{10}x - \log_x(0.01)$ for the domain $x > 1$.

At first glance, this looks like a messy collection of logarithms with different bases. However, in the context of JEE Advanced, such complexity is often a mask for underlying symmetry.

### Phase 1: The Simplification

Our first hurdle is the decimal $0.01$. In mathematics, decimals are often just fractions in disguise.

We know that $0.01 = \frac{1}{100} = 10^{-2}$. By substituting this into our second term, the expression becomes:

$$2 \log_{10}x - \log_x(10^{-2})$$

The number $10$ now appears in both terms. This is the intended path forward.

### Phase 2: The Power Rule and Base Change

Now, we invoke the power rule of logarithms: $\log_b(a^n) = n \log_b(a)$. Applying this to our second term, the exponent $-2$ moves to the front.

Since we are subtracting this term, we get $-(-2) \log_x(10)$, which simplifies to $+2 \log_x(10)$. Our expression is now:

$$2 \log_{10}x + 2 \log_x(10)$$

To unify the bases, we use the base change formula: $\log_x(10) = \frac{1}{\log_{10}x}$. Factoring out the $2$, we obtain:

$$2 \left( \log_{10}x + \frac{1}{\log_{10}x} \right)$$

### Phase 3: The Substitution and AM-GM

Let us define $t = \log_{10}x$. Because $x > 1$, we know that $t > 0$.

Our expression is now $y = 2(t + \frac{1}{t})$. We need the minimum value of $t + \frac{1}{t}$ for $t > 0$.

This is a classic application of the **Arithmetic Mean-Geometric Mean (AM-GM) inequality**. The inequality states that for positive numbers $a$ and $b$:

$$\frac{a+b}{2} \ge \sqrt{ab}$$

Setting $a = t$ and $b = \frac{1}{t}$, we get:

$$\frac{t + \frac{1}{t}}{2} \ge \sqrt{t \cdot \frac{1}{t}} = 1$$

This implies that $t + \frac{1}{t} \ge 2$.

### Final Calculation

Since $t + \frac{1}{t} \ge 2$, our expression $y = 2(t + \frac{1}{t})$ must satisfy:

$$y \ge 2(2) = 4$$

The **minimum value is 4**.

If the provided options do not include $4$, the correct choice is 'none of these'. Trusting your derivation is the hallmark of a successful engineer.

Animated Solution for Mathematics - Basic Mathematics: The least value of the expression $2 lo g_{10} x - lo g_{x} (0.01)$ , for $x > 1$ , is

Select Answer:

Visualized Solution

Understanding the Expression $2 lo g_{10} x - lo g_{x} (0.01)$

Simplifying $0.01$ as $1 0^{- 2}$

Applying the Power Rule

Using the Base Change Formula

Substituting $t = lo g_{10} x$

The $A M \geq GM$ Inequality

Calculating the Minimum Value

Final Conclusion

The Sigma Insight: Properties of Logarithms

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

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Animated Solution for Mathematics - Basic Mathematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least value of the expression 2log10​x−logx​(0.01), for x>1, is