对数与幂和问题：已知2^x+2^y=2^{f(x,y)}，求f(x,y)表达式

Hey there! Let’s walk through this problem clearly—you’re right that taking logs directly doesn’t work because we’re dealing with a sum of exponents, not a product. Here’s the step-by-step fix:

Step 1: Use symmetry to simplify the problem #

Since addition is commutative ((2^x + 2^y = 2^y + 2^x)), we can assume without loss of generality that (x \leq y) (we can swap (x) and (y) later if needed to cover all cases).

Step 2: Factor out the smaller power #

When (x \leq y), we can factor (2^x) out of the left-hand side to turn the sum into a product:

2^x + 2^y = 2^x(1 + 2^{y - x})

This product form is much easier to work with using logarithms.

Step 3: Take base-2 logarithms of both sides #

Since the right-hand side is (2^{f(x,y)}), taking (\log_2) of both sides gives us:

log₂(2^x(1 + 2^{y - x})) = f(x,y)

Using the logarithm product rule ((\log_b(ab) = \log_b a + \log_b b)), we split this into two simple terms:

f(x,y) = log₂(2^x) + log₂(1 + 2^{y - x})

We know (\log₂(2^x) = x), so this simplifies to:

f(x,y) = x + log₂(1 + 2^{y - x})

Step 4: Generalize for all (x) and (y) #

If (x > y), we just swap the roles of (x) and (y), leading to:

f(x,y) = y + log₂(1 + 2^{x - y})

We can write this in a unified form using the maximum function and absolute value to cover both cases neatly:

f(x,y) = max(x, y) + log₂(1 + 2^{-|x - y|})

Or even more concisely, since the original equation directly tells us (f(x,y)) is the base-2 logarithm of the sum:

f(x,y) = log₂(2^x + 2^y)

The max/absolute value form is often more useful for further analysis, while the direct log form is the most straightforward expression.

Quick verification #

Let’s test with (x = y):

• Left-hand side: (2^x + 2^x = 2 \cdot 2^x = 2^{x+1})
• Using our formula: (max(x,x) + log₂(1 + 2^0) = x + log₂(2) = x + 1), which matches perfectly.

Another test: (x=1, y=3):

• Left-hand side: (2 + 8 = 10 = 2^{log₂10})
• Formula gives (3 + log₂(1 + 2^{-2}) = 3 + log₂(5/4) = log₂(8) + log₂(5/4) = log₂(10)), which is correct.

内容的提问来源于stack exchange，提问作者R. Emery