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Algebra Level 1

2 0 + 2 − 1 2 − 2 + 2 − 3 = ? \large \dfrac{2^0 + 2^{-1}}{2^{-2} + 2^{-3}} = \, ?


The answer is 4.

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Gean Llego by Gean Llego , 25, Philippines

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5 solutions

Nihar Mahajan
Jan 13, 2016

2 0 + 2 − 1 2 − 2 + 2 − 3 = 2 2 ( 2 0 − 2 + 2 − 1 − 2 ) 2 − 2 + 2 − 3 = 2 2 ( 2 − 2 + 2 − 3 ) 2 − 2 + 2 − 3 = 2 2 = 4 \large \dfrac{2^0 + 2^{-1}}{2^{-2} + 2^{-3}} = \dfrac{2^2( \ 2^{0-2} + 2^{-1-2} \ )}{2^{-2} + 2^{-3}}=\dfrac{2^2( 2^{-2} + 2^{-3} )}{2^{-2} + 2^{-3}}= 2^2=\boxed{4}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice one mihar whats your age i did it the same

Adirta Puri - 5 years, 5 months ago
Gean Llego
Jan 13, 2016

We've done things manually.

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Multiplying numerator and denominator by 2³, the expression becomes:

(2³+2²)/(2+1)=(8+4)/3=12/3=4, the simplest method I got

3 Helpful 0 Interesting 0 Brilliant 0 Confused

2 0 + 2 − 1 2 − 2 + 2 − 3 \ = 1 + 1 2 1 4 + 1 8 = 1 + 1 2 3 8 = 24 2 3 = 12 3 ∴ 4 \begin{aligned} \frac{2^0+2^{-1}}{2^{-2}+2^{-3}}\ = \frac{1 + \frac{1}{2}}{ \frac{1}{4} + \frac{1}{8}} \\&= \frac{1+ \frac{1}{2}} {\frac{3}{8}} \\&= \frac{\frac{24}{2}}{3} \\& = \frac{12}{3}\\& \therefore \boxed{4} \end{aligned}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Abdul Hai
Jan 20, 2016

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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