What is the answer to 4^{x-1}+4^x+4^{x+1}=84 ?

The answer to 4^{x-1}+4^x+4^{x+1}=84 is x=2

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4^{x-1}+4^x+4^{x+1}=84

4^{x - 1} + 4^{x} + 4^{x + 1} = 84

x = 2

Solution steps

4^{x - 1} + 4^{x} + 4^{x + 1} = 84

Factor

4^{x - 1} + 4^{x} + 4^{x + 1} : 21 \cdot 4^{x - 1}

21 \cdot 4^{x - 1} = 84

Divide both sides by

21

4^{x - 1} = 4

Apply exponent rules

x - 1 = 1

Solve

x - 1 = 1 : x = 2

x = 2

x^{x} = 10

x^{x} = 27

9^{x} = 3^{x} - 4

k, n - 2 = 3^{k}

(2 7^{27})^{x} = (3^{27})^{17}

What is the answer to 4^{x-1}+4^x+4^{x+1}=84 ?
The answer to 4^{x-1}+4^x+4^{x+1}=84 is x=2
Solve for x: 4^{x-1}+4^x+4^{x+1}=84
The solution is x=2