What is ((2^{x+5}-4*2^x))/((16*2^{x+2)-2^{x+3})} ?

The solution to ((2^{x+5}-4*2^x))/((16*2^{x+2)-2^{x+3})} is 1/2

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simplify ((2^{x+5}-42^x))/((162^{x+2)-2^{x+3})}

simplify

\frac{( 2 ^{x + 5} - 4 \cdot 2 ^{x} )}{( 16 \cdot 2 ^{x + 2} - 2 ^{x + 3} )}

\frac{1}{2}

Solution steps

\frac{2 ^{x + 5} - 4 \cdot 2 ^{x}}{16 \cdot 2 ^{x + 2} - 2 ^{x + 3}}

Factor

2^{x + 5} - 4 \cdot 2^{x} : 2^{2 + x} \cdot 7

= \frac{2 ^{2 + x} \cdot 7}{16 \cdot 2 ^{x + 2} - 2 ^{x + 3}}

Factor

16 \cdot 2^{x + 2} - 2^{x + 3} : 2^{x + 3} \cdot 7

= \frac{2 ^{2 + x} \cdot 7}{2 ^{x + 3} \cdot 7}

Cancel the common factor:

7

= \frac{2 ^{2 + x}}{2 ^{x + 3}}

Simplify

\frac{2 ^{2 + x}}{2 ^{x + 3}} : \frac{1}{2}

= \frac{1}{2}

(a - 1) \cdot (a + 1) \cdot (a^{2} + a + 1) \cdot (a^{2} - a + 1)

2130829 k^{1} - 18041807671

3 sin^{2} (a) + 3 cos^{2} (a)

36 p^{2} + 84 p + 49

(x - 1) (\frac{x - 5}{4}) (\frac{x + 1}{4})

What is ((2^{x+5}-4*2^x))/((16*2^{x+2)-2^{x+3})} ?
The solution to ((2^{x+5}-4*2^x))/((16*2^{x+2)-2^{x+3})} is 1/2