Let f(x) = 2^x + 3^x
Since 2^x and 3^x are strictly increasing functions, f(x) is a strictly increasing function.
That is to say, if x < y, then f(x) < f(y).
Strictly increasing functions are one to one. If f(x) = f(y), then x = y,
So, f(x) = 13 can have at most one solution.
f(0) = 2
f(1) = 5
f(2) = 13
So, we see that x = 2 is a solution.
Hence x = 2 is the only real valued solution.
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