Question

Let $f(x) = x^3$ and $g(x) = 3^x$. The values of $a$ such that $g(f(a)) = f(g(a))$ are:

• A. $0, 2$
• B. $1, 3$
• C. $0, \pm 3$
• D. $1, \pm 2$
• E. $0, \pm \sqrt{3}$

Hint:

Remember to use the law of exponents $(x^m)^n = x^{mn}$ when calculating $f(g(a))$. A common mistake is to write $3^{a^3}$ for both compositions, but the placement of the power matters significantly in exponentiation.

Answer 1

The Correct Option is C

Concept: This problem involves composite functions and solving an equation where the order of composition is swapped. We must find the expressions for $g(f(a))$ and $f(g(a))$ and set them equal to each other.

Step 1: Find the expressions for the composite functions.
For $g(f(a))$: Substitute $f(a) = a^3$ into $g(x) = 3^x$. \[ g(f(a)) = 3^{(a^3)} \] For $f(g(a))$: Substitute $g(a) = 3^a$ into $f(x) = x^3$. \[ f(g(a)) = (3^a)^3 = 3^{3a} \]

Step 2: Equate the two expressions.
Set $g(f(a)) = f(g(a))$: \[ 3^{a^3} = 3^{3a} \] Since the bases are the same, we can equate the exponents: \[ a^3 = 3a \]

Step 3: Solve the resulting equation for $a$.
Rearrange to solve the cubic equation: \[ a^3 - 3a = 0 \] Factor out the common term $a$: \[ a(a^2 - 3) = 0 \] The solutions are:
• $a = 0$
• $a^2 - 3 = 0 \Rightarrow a^2 = 3 \Rightarrow a = \pm \sqrt{3}$ The values of $a$ are $0, \pm \sqrt{3}$.