What is the answer to solvefor x,4^{x+1}+2^y=20 ?

The answer to solvefor x,4^{x+1}+2^y=20 is x=(ln(20-2^y))/(2ln(2))-1

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solvefor x,4^{x+1}+2^y=20

solve for

x, 4^{x + 1} + 2^{y} = 20

x = \frac{ln ( 20 - 2 ^{y} )}{2 ln ( 2 )} - 1

Solution steps

4^{x + 1} + 2^{y} = 20

Move

2^{y}

to the right side

4^{x + 1} = 20 - 2^{y}

Apply exponent rules

2 (x + 1) ln (2) = ln (20 - 2^{y})

Solve

2 (x + 1) ln (2) = ln (20 - 2^{y}) : x = \frac{ln ( 20 - 2 ^{y} )}{2 ln ( 2 )} - 1

x = \frac{ln ( 20 - 2 ^{y} )}{2 ln ( 2 )} - 1

t, (1.045)^{t} = 2

x, 1 0^{- 6 x} = 10 0^{3 - 2 x}

3 \cdot 2^{x + 1} - 64 = 2^{x + 1}

(5)^{x} = 125

What is the answer to solvefor x,4^{x+1}+2^y=20 ?
The answer to solvefor x,4^{x+1}+2^y=20 is x=(ln(20-2^y))/(2ln(2))-1