f(xy) = f(x) + f(y) is true for all
Logarithmic function f
The question asks to identify the type of function \(f\) that satisfies the functional equation \(f(xy) = f(x) + f(y)\) for all valid inputs \(x\) and \(y\). This equation is a specific property that only certain types of functions fulfill.
The equation \(f(xy) = f(x) + f(y)\) relates the function's value at the product of two variables to the sum of its values at each individual variable. We need to examine the given types of functions to see which one exhibits this property.
A general polynomial function is of the form \(f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\). Let's consider a simple non-constant example, say \(f(x) = x^2\).
Generally, \(x^2 y^2 \neq x^2 + y^2\) (for example, if \(x=2, y=3\), \(f(xy) = (6)^2 = 36\), but \(f(x) + f(y) = 2^2 + 3^2 = 4 + 9 = 13\)). This property does not hold for polynomial functions in general (except for the trivial case \(f(x) = 0\), which is a constant polynomial, but the property usually refers to non-trivial cases or holds for specific domains where other functions are defined).
Consider a common trigonometric function like \(f(x) = \sin(x)\).
In general, \(\sin(xy) \neq \sin(x) + \sin(y)\) (for example, if \(x = \pi/2, y=1\), \(f(xy) = \sin(\pi/2) = 1\), but \(f(x) + f(y) = \sin(\pi/2) + \sin(1) = 1 + \sin(1)\), which is not 1). Trigonometric functions do not satisfy this functional equation.
Consider an exponential function \(f(x) = a^x\) where \(a > 0\) and \(a \neq 1\).
In general, \(a^{xy} \neq a^x + a^y\) (for example, if \(a=2, x=2, y=3\), \(f(xy) = 2^{6} = 64\), but \(f(x) + f(y) = 2^2 + 2^3 = 4 + 8 = 12\)). Exponential functions do not satisfy this functional equation.
Consider a logarithmic function \(f(x) = \log_b(x)\) where \(b > 0\) and \(b \neq 1\), and typically \(x > 0, y > 0\) for the logarithms to be defined for real numbers.
A fundamental property of logarithms states that the logarithm of a product is the sum of the logarithms: \(\log_b(xy) = \log_b(x) + \log_b(y)\). Therefore, the logarithmic function \(f(x) = \log_b(x)\) directly satisfies the functional equation \(f(xy) = f(x) + f(y)\).
Based on our analysis, the functional equation \(f(xy) = f(x) + f(y)\) is a defining characteristic of logarithmic functions.
| Function Type | General Form Example | Check f(xy) = f(x) + f(y) | Satisfies Property? |
|---|---|---|---|
| Polynomial | \(f(x) = x^n\) | \(x^n y^n\) vs \(x^n + y^n\) | No (generally) |
| Trigonometric | \(f(x) = \sin(x)\) | \(\sin(xy)\) vs \(\sin(x) + \sin(y)\) | No (generally) |
| Exponential | \(f(x) = a^x\) | \(a^{xy}\) vs \(a^x + a^y\) | No (generally) |
| Logarithmic | \(f(x) = \log_b(x)\) | \(\log_b(xy) = \log_b(x) + \log_b(y)\) | Yes |
Therefore, the functional equation \(f(xy) = f(x) + f(y)\) is true for logarithmic functions.
Here's a quick review of key properties for different function types:
The equation \(f(xy) = f(x) + f(y)\) is a famous functional equation. It is a variant of Cauchy's functional equations. For continuous functions defined on positive real numbers, the solutions are of the form \(f(x) = c \log_b(x)\) for some constant \(c\), or equivalently \(f(x) = C \ln(x)\) for some constant \(C\), where \(\ln\) is the natural logarithm. If the domain includes zero or negative numbers, the definition and properties of the function become more complex.
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