Find the smallest value for $\frac1x+\frac2y$ if $\log x+\log y=2$

We are tasked with finding the smallest value of the expression $\frac{1}{x} + \frac{2}{y}$ subject to the constraint $\log x + \log y = 2$.

### Step 1: Simplify the constraint

First, simplify the given constraint:
\[
\log x + \log y = \log(xy)
\]
Thus, we have:
\[
\log(xy) = 2
\]
Exponentiating both sides, we obtain:
\[
xy = 10^2 = 100
\]

### Step 2: Express the objective function
We need to minimize the expression:
\[
\frac{1}{x} + \frac{2}{y}
\]
Given that $xy = 100$, we can express $y$ in terms of $x$ as:
\[
y = \frac{100}{x}
\]
Substituting $y = \frac{100}{x}$ into the objective function, we have:
\[
\frac{1}{x} + \frac{2}{y} = \frac{1}{x} + \frac{2x}{100} = \frac{1}{x} + \frac{x}{50}
\]
Thus, we need to minimize the function:
\[
f(x) = \frac{1}{x} + \frac{x}{50}
\]

### Step 3: Find the critical points
To find the minimum, we take the derivative of $f(x)$ with respect to $x$ and set it equal to zero:
\[
f'(x) = -\frac{1}{x^2} + \frac{1}{50}
\]
Setting $f'(x) = 0$ gives:
\[
-\frac{1}{x^2} + \frac{1}{50} = 0
\]
\[
\frac{1}{x^2} = \frac{1}{50}
\]
\[
x^2 = 50
\]
\[
x = \sqrt{50} = 5\sqrt{2}
\]

### Step 4: Confirm the minimum
We need to confirm that this critical point corresponds to a minimum. The second derivative of $f(x)$ is:
\[
f''(x) = \frac{2}{x^3}
\]
Since $f''(x) > 0$ for all $x > 0$, $f(x)$ is concave up, confirming that $x = 5\sqrt{2}$ is a minimum.

### Step 5: Calculate the minimum value
Substitute $x = 5\sqrt{2}$ back into the expression for $f(x)$:
\[
f(5\sqrt{2}) = \frac{1}{5\sqrt{2}} + \frac{5\sqrt{2}}{50}
\]
Simplify each term:
\[
\frac{1}{5\sqrt{2}} = \frac{\sqrt{2}}{10}
\]
\[
\frac{5\sqrt{2}}{50} = \frac{\sqrt{2}}{10}
\]
Thus:
\[
f(5\sqrt{2}) = \frac{\sqrt{2}}{10} + \frac{\sqrt{2}}{10} = \frac{2\sqrt{2}}{10} = \frac{\sqrt{2}}{5}
\]

### Final Answer
The smallest value of $\frac{1}{x} + \frac{2}{y}$ is $\boxed{\frac{\sqrt{2}}{5}}$.

Would you like more details or have any questions about this solution?

Here are 5 related questions:
1. What would happen if we changed the constraint to $\log x + 2\log y = 2$?
2. How does the AM-GM inequality apply to problems like this?
3. What is the geometric interpretation of minimizing the expression under the given constraint?
4. How would the solution change if the constraint were $x \cdot y = k$ for a general constant $k$?
5. Can this problem be solved using Lagrange multipliers?

**Tip:** When minimizing or maximizing functions with constraints, expressing one variable in terms of another often simplifies the problem significantly.

Explore how to minimize the expression 1/x + 2/y under the constraint log x + log y = 2. Learn the step-by-step solution involving logarithm rules and optimization techniques. Suitable for advanced high school students and those preparing for mathematical competitions.

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