Solve the equation 1+3^(x ∕ 2)=2^x.

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We'll bring the equation to the form:

(1/2)^x + {[(3)^1/2]/2}^x=1

If we'll put x=2:

(1/2)^2 + {[(3)^1/2]/2}^2=1

After inputting the x=2, we'll see that the equation is verified.

1/4 + 3/4 =4/4 =1 true

Now, all we have to do is to prove that the x=2 is the unique solution for the equation above.

We'll consider the expression (1/2)^x + {[(3)^1/2]/2}^x as being the law of the function f(x)= (1/2)^x + {[(3)^1/2]/2}^x

After a short analysis we'll notice that f(x) is a sum of exponential functions whose base are subunits.

It is known that an exponential function whose base is subunit is a decreasing function. Based on this, for x=1<x=2, we'll have f(1)>f(2)=1.

For x=3>x=2, we'll have f(3)<f(2)=1.

In summary, any real number x, less or greater than x= 2, could not be the root of the equation 1+3^(x ∕ 2)=2^x.

**So the root x=2, is unique.**

1+3^(x/2) = 2^x. Or

3^(x/2) = 2^x-1. Squaring both sides,

3^x = (2^x)^2 -2*2^x+1. Or

f(x) =4^x-3^x-2*2^x+1 = 0.

For f(x) is -ve for x= 0 and 1 and

For x =2 , LHs = 3^2 = 9 and RHS =4^2-2*2^2+1 = 9.

So x =2 is the solution.

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