What is the answer to 4^{x+1}+4^{x+2}=1280 ?

The answer to 4^{x+1}+4^{x+2}=1280 is x=3

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4^{x+1}+4^{x+2}=1280

4^{x + 1} + 4^{x + 2} = 1280

x = 3

Solution steps

4^{x + 1} + 4^{x + 2} = 1280

Factor

4^{x + 1} + 4^{x + 2} : 5 \cdot 4^{x + 1}

5 \cdot 4^{x + 1} = 1280

Divide both sides by

5

4^{x + 1} = 256

Apply exponent rules

x + 1 = 4

Solve

x + 1 = 4 : x = 3

x = 3

3^{x + 2} = \frac{1}{81}

3^{2 x} = 7^{3 x - 1}

9 \cdot 5^{- x} + 5^{x} = 10

8^{9 x - 2} = 64

2^{2 x} = 4^{x}

What is the answer to 4^{x+1}+4^{x+2}=1280 ?
The answer to 4^{x+1}+4^{x+2}=1280 is x=3
Solve for x: 4^{x+1}+4^{x+2}=1280
The solution is x=3