What is the answer to 2^x+1/(2^{x-2)}=5 ?

The answer to 2^x+1/(2^{x-2)}=5 is x=2,x=0

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2^x+1/(2^{x-2)}=5

2^{x} + \frac{1}{2 ^{x - 2}} = 5

x = 2, x = 0

Solution steps

2^{x} + \frac{1}{2 ^{x - 2}} = 5

Apply exponent rules

2^{x} + (2^{x})^{- 1} \cdot 2^{2} = 5

Rewrite the equation with

2^{x} = u

u + (u)^{- 1} \cdot 2^{2} = 5

Solve

u + u^{- 1} \cdot 2^{2} = 5 : u = 4, u = 1

u = 4, u = 1

Substitute back

u = 2^{x},

solve for

x

Solve

2^{x} = 4 : x = 2

Solve

2^{x} = 1 : x = 0

x = 2, x = 0

Verify Solutions:

x = 2

True

, x = 0

True

The solutions are

x = 2, x = 0

14 \cdot 4^{x - 1} = 2 \cdot 7^{x - 1}

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

4^{x} - 2^{x} = 6

x, y = 4^{x} y

x, e^{x} = 2^{y}

What is the answer to 2^x+1/(2^{x-2)}=5 ?
The answer to 2^x+1/(2^{x-2)}=5 is x=2,x=0
Solve for x: 2^x+1/(2^{x-2)}=5
The solution is x=2,x=0