What is the answer to 4^{x+1}+4^{x+2}=128 ?

The answer to 4^{x+1}+4^{x+2}=128 is x=(ln(128/5))/(2ln(2))-1

Solve for x: 4^{x+1}+4^{x+2}=128

The solution is x=(ln(128/5))/(2ln(2))-1

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4^{x + 1} + 4^{x + 2} = 128

x = \frac{ln ( \frac{128}{5} )}{2 ln ( 2 )} - 1

Decimal Notation

x = 1.33903 \dots

Solution steps

4^{x + 1} + 4^{x + 2} = 128

Factor

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

5 \cdot 4^{x + 1} = 128

Divide both sides by

5

4^{x + 1} = \frac{128}{5}

Apply exponent rules

(x + 1) ln (4) = ln (\frac{128}{5})

Solve

(x + 1) ln (4) = ln (\frac{128}{5}) : x = \frac{ln ( \frac{128}{5} )}{2 ln ( 2 )} - 1

x = \frac{ln ( \frac{128}{5} )}{2 ln ( 2 )} - 1

8 e^{4 x} - 32 e^{- 4 x} = 0

3^{2 x} = 2 e^{x}

2^{n} = 10

2^{3 x} - 2^{x} = 120

4 9^{x} - 6 \cdot 7^{x} - 7 = 0

What is the answer to 4^{x+1}+4^{x+2}=128 ?
The answer to 4^{x+1}+4^{x+2}=128 is x=(ln(128/5))/(2ln(2))-1
Solve for x: 4^{x+1}+4^{x+2}=128
The solution is x=(ln(128/5))/(2ln(2))-1