What is the answer to 4^{sqrt(x+1)}-2^{sqrt(x+1)+2}=0 ?

The answer to 4^{sqrt(x+1)}-2^{sqrt(x+1)+2}=0 is x=3

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4^{sqrt(x+1)}-2^{sqrt(x+1)+2}=0

4^{x + 1} - 2^{x + 1 + 2} = 0

x = 3

Solution steps

4^{x + 1} - 2^{x + 1 + 2} = 0

Add

2^{x + 1 + 2}

to both sides

4^{x + 1} - 2^{x + 1 + 2} + 2^{x + 1 + 2} = 0 + 2^{x + 1 + 2}

Simplify

4^{x + 1} = 2^{x + 1 + 2}

Apply exponent rules

2 x + 1 = x + 1 + 2

Solve

2 x + 1 = x + 1 + 2 : x = 3

x = 3

x^{x} = \frac{2}{2}

25 6^{3 x + 3} = 102 4^{2 x + 2}

216 = 6^{2 r - 11}

2^{x - x^{2}} = \frac{1}{4 ^{x}}

62 5^{x} = \frac{1}{5}

What is the answer to 4^{sqrt(x+1)}-2^{sqrt(x+1)+2}=0 ?
The answer to 4^{sqrt(x+1)}-2^{sqrt(x+1)+2}=0 is x=3
Solve for x: 4^{sqrt(x+1)}-2^{sqrt(x+1)+2}=0
The solution is x=3