We notice that the value x = 1 is the solution of the equation.

2^1 + 3^1 = 5^1

2 + 3 = 5

Now, we'll have to verify if x = 1 is the only solution for the given equation or if there are more.

We'll divide the equation, both sides by 5^x:

(2/5)^x + (3/5)^x = 1

We'll assign a function f(x) to the expresison (2/5)^x + (3/5)^x.

The exponential functions (2/5)^x and (3/5)^x are decreasing functions (the denominator is bigger than numerator), so f(x) is a decreasing function, too.

If f(x) is a decreasing function, it could have only one solution x = 1.

(2/5)^1 + (3/5)^1 = 1

(2+3)/5 = 1

5/5 = 1

1 = 1

**So, x = 1 is the only solution for the equation:**

** 2^x + 3^x = 5^x**

2^x+3^x = 5^x.

Let f(x) = 2^x+3^x - 5^x.

We see that f(0) = 2^0+2^0 -5^0 = 1+2-1 = 1.

f(1) = 2^1+3^1-5^1 = 2+3-5 = 0 . So x= 1 is a solution.

f(1+h) = -ve .

Also f(x) < 0 for x>1.

Also f(x) > 0 for all x <1.

Therefore there is only one solution for 2^x+3^x=5^x when x= 1.