Problem : 1 The least value of the expression $$ 2 \log _{10} x-\log _{x}(0.01) $$, for $$ x1 $$, is [IIT 1980 $$ \begin{array}{ll} ext { (a) } 10 & ext { (b) } 2 \ ext { (c) }-0.01 & ext { (d) None of these }\end{array} $$

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Problem : 1 The least value of the expression $$ 2 \log _{10} x-\log _{x}(0.01) $$, for $$ x>1 $$, is [IIT 1980 $$ \begin{array}{ll}	ext { (a) } 10 & 	ext { (b) } 2 \ 	ext { (c) }-0.01 & 	ext { (d) None of these }\end{array} $$

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To find the least value of the expression $$ 2 \log_{10} x - \log_x(0.01) $$ for $$ x > 1 $$, we can proceed as follows:

1. **Introduce a Substitution:**
   Let $$ y = \log_{10} x $$. Since $$ x > 1 $$, it follows that $$ y > 0 $$.

2. **Rewrite the Expression:**
   The expression becomes:
   $$
   2y - \log_x(0.01)
   $$
   
   Note that $$ 0.01 = 10^{-2} $$, so:
   $$
   \log_x(0.01) = \log_x(10^{-2}) = -2 \log_x(10)
   $$
   
   Therefore, the expression simplifies to:
   $$
   2y + 2 \log_x(10)
   $$
   
   Using the change of base formula:
   $$
   \log_x(10) = \frac{1}{\log_{10} x} = \frac{1}{y}
   $$
   
   So, the expression now is:
   $$
   2y + \frac{2}{y}
   $$

3. **Optimize the Expression:**
   To minimize $$ 2y + \frac{2}{y} $$, take the derivative with respect to $$ y $$ and set it to zero:
   $$
   f(y) = 2y + \frac{2}{y}
   $$
   $$
   f'(y) = 2 - \frac{2}{y^2}
   $$
   Setting $$ f'(y) = 0 $$:
   $$
   2 - \frac{2}{y^2} = 0 \implies y^2 = 1 \implies y = 1 \quad (\text{since } y > 0)
   $$
   
   The minimum value is then:
   $$
   f(1) = 2(1) + \frac{2}{1} = 4
   $$

4. **Conclusion:**
   The minimum value of the expression is **4**, which is not among the provided options.

**Answer:**  
Option (d) None of these
The least value of the expression is 4, which is not among the given options.

Problem : 1
The least value of the expression , for , is
[IIT 1980

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Problem : 1 The least value of the expression , for , is [IIT 1980