if-3-x-24-and-2-y-36-find-4-x-1-y-4-x-

Question Number 145548 by mathdanisur last updated on 05/Jul/21

i f 3^{x} = 24 a n d 2^{y} = 36

f i n d \frac{4^{(x - 1) \cdot y}}{4^{x}} = ?

Answered by Ar Brandon last updated on 06/Jul/21

\frac{4^{(x - 1) y}}{4^{x}} = \frac{2^{2 y (x - 1)}}{4^{x}} = \frac{36^{2 (x - 1)}}{4^{x}} = \frac{{(4 \times 9)}^{2 x - 2}}{4^{x}}

= 4^{x - 2} \times 3^{4 x - 4} = \frac{4^{x}}{16} \times \frac{3^{4 x}}{81} = \frac{4^{x}}{16} \times \frac{24^{4}}{81}

= 4^{\log_{3} 24} \times 2^{8} = 2^{{2 \log}_{3} 24 + 8}

Answered by imjagoll last updated on 06/Jul/21

from 2^{y} = 36 & 3^{x} = 24

\Rightarrow {(3^{x - 1})}^{y} = {(2^{3})}^{y}

\Rightarrow 3^{xy - y} = 36^{3} \Rightarrow xy - y = \log_{3} {(36)}^{3}

\Rightarrow 4^{xy - y} = 4^{\log_{3} {(36)}^{3}}

\Rightarrow 4^{x} = 4^{\log_{3} (24)}

\Rightarrow \frac{4^{xy - y}}{4^{x}} = 4^{\log_{3} (\frac{36^{3}}{24})} = 4^{\log_{3} (1944)}

\Rightarrow 1944^{\log_{3} (4)} = {(2^{3} \times 3^{5})}^{\log_{3} (4)}

= 2^{3 . \log_{3} (4)} \times 3^{5 . \log_{3} (4)}

= 4^{5} \times 2^{6 . \log_{3} (2)} \approx 14, 121.233767

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