If f(x) = (3/5)^x + (4/5)^x, x ∈ R, then the equation f(x) = 0 has how many solutions?

f(x) = (3/5)^x + (4/5)^x
Substituting sinθ = 3/5 and cosθ = 4/5, we get
f(x) = sinθ^x + cosθ^x
Now for equation f(x) = 0
⇒ sinθ^x + cosθ^x = 0
⇒ sinθ^x = -cosθ^x
Squaring both sides,
sin²θ^x = cos²θ^x
⇒ sin²θ^x = 1 - sin²θ^x
⇒ 2sin²θ^x = 1
⇒ sin²θ^x = 1/2
⇒ sinθ^x = ±1/√2
This implies that θ^x = π/4 or θ^x = 3π/4
However, this is incorrect as the sum of two positive terms cannot be 0.
Let's analyze the function f(x) = (3/5)^x + (4/5)^x.
Both terms are always positive for any real x. Therefore, their sum is always positive and can never be equal to 0.
Thus, there is no solution to the equation f(x) = 0.