What is the answer to 2^{x+1}+2^{2-x}=9 ?

The answer to 2^{x+1}+2^{2-x}=9 is x=2,x=-1

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2^{x+1}+2^{2-x}=9

2^{x + 1} + 2^{2 - x} = 9

x = 2, x = - 1

Solution steps

2^{x + 1} + 2^{2 - x} = 9

Apply exponent rules

2^{x} \cdot 2^{1} + (2^{x})^{- 1} \cdot 2^{2} = 9

Rewrite the equation with

2^{x} = u

u \cdot 2^{1} + (u)^{- 1} \cdot 2^{2} = 9

Solve

u \cdot 2^{1} + u^{- 1} \cdot 2^{2} = 9 : u = 4, u = \frac{1}{2}

u = 4, u = \frac{1}{2}

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 4 : x = 2

Solve

2^{x} = \frac{1}{2} : x = - 1

x = 2, x = - 1

3 = 1 0^{t}

4^{3 x - 1} = \frac{1}{8 ^{x - 1}}

r, y = r (x - 1)^{r} + 1

2^{x^{2}} - 4^{3 x - 4} = 0

What is the answer to 2^{x+1}+2^{2-x}=9 ?
The answer to 2^{x+1}+2^{2-x}=9 is x=2,x=-1
Solve for x: 2^{x+1}+2^{2-x}=9
The solution is x=2,x=-1