Fractions with Exponents

Algebra Level 2

Solve the equation: 2 2 x + 1 + 2 2 x = 3 8 . \large 2^{2x+1} +2^{2x} = \dfrac{3}{8 }.

2 3 \dfrac{2}{3} − 3 2 \dfrac{-3}{2} 2 2 None of these 3 2 \dfrac{3}{2}

Cooper Haddrill by Cooper Haddrill , 21, USA

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2 solutions

● 2 2 x + 1 + 2 2 x = 3 8 2^{2x+1} +2^{2x} =\frac{3}{8 }
2 2 x × 2 1 + 2 2 x = 3 8 2^{2x}×2^1+2^{2x}=\frac{3}{8}
→Let 2 2 x = y 2^{2x}=y .
y × 2 + y = 3 8 y×2+y=\frac{3}{8}
3 y = 3 8 3y=\frac{3}{8}
y = 1 8 y=\frac{1}{8}
→Now, substitute y = 2 2 x y=2^{2x} .
2 2 x = 2 − 3 2^{2x}=2^{-3}
2 x = − 3 2x=-3
x = − 3 2 x=\boxed{\frac{-3}{2}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Kay Xspre
Nov 15, 2015

Factorize it gives 2 2 x ( 2 + 1 ) = 3 ( 2 − 3 ) 2^{2x}(2+1) = 3(2^{-3}) , or simply 2 x = − 3 ; x = − 3 2 2x = -3;\: x = -\frac{3}{2}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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