ANSWER
[tex]f(x) = 14x - 5[/tex]
EXPLANATION
The given expression is
[tex]( \frac{1}{32})^{x} \times ( \frac{1}{2})^{9x - 5} = ( \frac{1}{2} ) ^{f(x)} [/tex]
We rewrite the left hand side using the laws of indices.
[tex]( \frac{1}{2})^{5x} \times ( \frac{1}{2})^{9x - 5} = ( \frac{1}{2} ) ^{f(x)} [/tex]
The bases are now the same on the LHS.
We write one base and add the exponents
[tex] ( \frac{1}{2})^{5x + 9x - 5} = ( \frac{1}{2} ) ^{f(x)} [/tex]
[tex] ( \frac{1}{2})^{14x- 5} = ( \frac{1}{2} ) ^{f(x)} [/tex]
Since the bases are equal, the exponents are also equal:
[tex]f(x) = 14x - 5[/tex]
