If-2-x-4-y-8-z-and-xyz-288-then-find-1-2x-1-4y-1-8z-

Question Number 113876 by Aina Samuel Temidayo last updated on 16/Sep/20

If 2^{x} = 4^{y} = 8^{z} and xyz = 288, then find

\frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z}

Answered by Olaf last updated on 16/Sep/20

2^{x} = 4^{y} = 8^{z}

\log_{2} 2^{x} = \log_{2} 4^{y} = \log_{2} 8^{z}

x \log_{2} 2 = y \log_{2} 2^{2} = z \log_{2} 2^{3}

x = 2 y = 3 z

x y z = 288

x (\frac{x}{2}) (\frac{x}{3}) = 288

x^{3} = 6 \times 288 = 1728

x = \overset{3}{} \sqrt{1728} = \overset{3}{} \sqrt{12^{3}} = 12

y = \frac{x}{2} = 6

z = \frac{x}{3} = 4

\frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z} = \frac{1}{2.12} + \frac{1}{4.6} + \frac{1}{8.4}

= \frac{1}{24} + \frac{1}{24} + \frac{1}{32} = \frac{11}{96}

Commented by Aina Samuel Temidayo last updated on 16/Sep/20

Thanks .

Answered by MJS_new last updated on 16/Sep/20

2^{x} = {(2^{2})}^{y} = {(2^{3})}^{z} \Rightarrow y = \frac{x}{2} \land z = \frac{x}{3}

x y z = \frac{x^{3}}{6} = 288 \Rightarrow x = 12

\frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z} = \frac{11}{8 x} = \frac{11}{96}

Answered by john santu last updated on 16/Sep/20

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

(2) 2^{x} = 2^{3 z} \Rightarrow x = 3 z

(3) 2^{2 y} = 2^{3 z} \Rightarrow y = \frac{3 z}{2}

(4) x y z = 288 \Rightarrow x (\frac{x}{2}) (\frac{x}{3}) = 12^{2} . 2

\Rightarrow x^{3} = 12^{3} \to {\begin{cases} x = 12 \\ y = 6 \\ z = 4 \end{cases}

t h e r e f o r e \frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z} = \frac{1}{24} + \frac{1}{24} + \frac{1}{32}

= \frac{2}{24} + \frac{1}{32} = \frac{2}{8.3} + \frac{1}{8.4} = \frac{8 + 3}{8.3.4} = \frac{11}{96}

Commented by Aina Samuel Temidayo last updated on 16/Sep/20

Thanks .

Answered by floor(10²Eta[1]) last updated on 16/Sep/20

x = 2 y = 3 z

xyz = 3 z . \frac{3 z}{2} . z = \frac{{9 z}^{3}}{2} = 288 \Rightarrow z = 4

\Rightarrow x = 12 and y = 6

\frac{1}{2 x} + \frac{1}{4 y} + \frac{1}{8 z} = \frac{1}{24} + \frac{1}{24} + \frac{1}{32} = \frac{11}{96}