What is the answer to solvefor q,2^x+1=sqr3(1-2^{x-1}) ?

The answer to solvefor q,2^x+1=sqr3(1-2^{x-1}) is q=(2^x+1)/(sr^3(-2^{-1+x)+1)};\quad x\ne 1

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solvefor q,2^x+1=sqr3(1-2^{x-1})

solve for

q, 2^{x} + 1 = s q r^{3} (1 - 2^{x - 1})

q = \frac{2 ^{x} + 1}{s r ^{3} ( - 2 ^{- 1 + x} + 1 )}; x \neq = 1

Solution steps

2^{x} + 1 = s q r^{3} (1 - 2^{x - 1})

Switch sides

s q r^{3} (1 - 2^{x - 1}) = 2^{x} + 1

Divide both sides by

s r^{3} (1 - 2^{x - 1}); x \neq = 1

q = \frac{2 ^{x} + 1}{s r ^{3} ( - 2 ^{- 1 + x} + 1 )}; x \neq = 1

3 x^{2} - 19 x - 14 = 0

- - 49

145

\frac{8}{6}

13 \leq 2 y + 17 \leq 21

What is the answer to solvefor q,2^x+1=sqr3(1-2^{x-1}) ?
The answer to solvefor q,2^x+1=sqr3(1-2^{x-1}) is q=(2^x+1)/(sr^3(-2^{-1+x)+1)};\quad x\ne 1