Find the smallest value for $\frac1x+\frac1y$ if $\log x+\log y=2$

We need to find the smallest value of the expression $\frac{1}{x} + \frac{1}{y}$ given the constraint $\log x + \log y = 2$.

### Step 1: Simplify the constraint

First, let's simplify the constraint:
\[
\log x + \log y = \log(xy) = 2
\]
Exponentiating both sides, we get:
\[
xy = 10^2 = 100
\]

### Step 2: Express the objective function

We want to minimize the expression:
\[
\frac{1}{x} + \frac{1}{y}
\]
Given the constraint $xy = 100$, we can express $y$ in terms of $x$ as:
\[
y = \frac{100}{x}
\]
Substituting $y = \frac{100}{x}$ into the objective function, we get:
\[
\frac{1}{x} + \frac{1}{y} = \frac{1}{x} + \frac{x}{100}
\]
So, we need to minimize the function:
\[
f(x) = \frac{1}{x} + \frac{x}{100}
\]

### 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:
\[
f'(x) = -\frac{1}{x^2} + \frac{1}{100}
\]
Setting $f'(x) = 0$ gives:
\[
-\frac{1}{x^2} + \frac{1}{100} = 0
\]
\[
\frac{1}{x^2} = \frac{1}{100}
\]
\[
x^2 = 100
\]
\[
x = \sqrt{100} = 10
\]

### Step 4: Confirm the minimum

To ensure that this critical point corresponds to a minimum, we examine the second derivative of $f(x)$:
\[
f''(x) = \frac{2}{x^3}
\]
Since $f''(x) > 0$ for all $x > 0$, the function $f(x)$ is concave up, confirming that $x = 10$ is indeed a minimum.

### Step 5: Calculate the minimum value

Substitute $x = 10$ back into the expression for $f(x)$:
\[
f(10) = \frac{1}{10} + \frac{10}{100} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}
\]

### Final Answer

The smallest value of $\frac{1}{x} + \frac{1}{y}$ is $\boxed{\frac{1}{5}}$.

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

Here are 5 related questions:
1. How would the problem change if the constraint was $\log x + \log y = 1$?
2. Can the AM-GM inequality provide an alternate approach to solving this problem?
3. What happens to the minimum value if $xy = k$, where $k$ is any positive constant?
4. How does the choice of base for the logarithm affect the solution?
5. Can Lagrange multipliers be used to solve this problem more generally?

**Tip:** When minimizing sums of reciprocals, it's often helpful to look for symmetries or constraints that allow simplification, such as expressing one variable in terms of another.

Discover how to minimize the expression 1/x + 1/y given the logarithmic constraint log x + log y = 2. This problem involves logarithm properties, derivative rules, and techniques for minimizing functions. Suitable for advanced high school students.

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