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

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

Solve for x: 4^x+4^{x-1}=4

The solution is x=(ln(4/5))/(2ln(2))+1

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

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

Decimal Notation

x = 0.83903 \dots

Solution steps

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

Factor

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

5 \cdot 4^{x - 1} = 4

Divide both sides by

5

4^{x - 1} = \frac{4}{5}

Apply exponent rules

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

Solve

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

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

2 (2^{2 x}) - 6 (2^{x}) + 4 = 0

12 = 200 (1 - e^{- \frac{t}{800}})

3^{(x + 1)} = 2^{3 x}

e^{l n (6 x)} = 19

(\frac{3}{4})^{x} = \frac{16}{9}

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