Is logarithms required?

Algebra Level 3

If $2^{x}=4^{y}=8^{z}$ and $xyz= 288$ , with $\frac{1}{2x} + \frac{1}{4y}+ \frac{1}{8z}= \frac{m}{n}$ for coprime positive integers $m$ and $n$ , find $m+n$ .

The answer is 107.

5 solutions

Sakanksha Deo
Feb 28, 2015

Simple,

Easily can be seen that

$x = 2y = 3z$

Therefore,

$6xyz = x^3$

Therefore

$x = 12$

The rest is too easy to explain.

Caleb Townsend
Feb 5, 2015

Since $2^x = 4^y = 8^z$ , $y = \frac{x}{\log_2(4)} = \frac{x}{2}, \ z = \frac{x}{\log_2(8)} = \frac{x}{3}$ and since $xyz = 288$ , substituting the above equations and solving for $x$ gives $x = \sqrt[3]{1728} = 12.$ Finally, $\frac{1}{24} + \frac{1}{24} + \frac{1}{32} = \frac{11}{96}$ $11 + 96 = \boxed{107}$

Oh! I didn't see that it was 1/(2x) + 1/(4y) + 1/(8z) ! I thought I was supposed to figure out what m and n were after summing (12/2 + 6/4 + 4/8). I didn't think that was possible, especially since m and n were not necessarily prime. I wish this problem had been easier to read. :p

Jared Jones - 6 years, 4 months ago

Sai Ram
Jul 13, 2015

Rest can be easily calculated.

Bala Vidyadharan
Jul 5, 2015

By making the bases as same we get x=2y=3z

let x=2y=3z=k

x=k ,y=k/2,z=k/3

sub the above values in xyz=288

k.k/2.k/3=288

k^3/6=288

k^3=1728

k=12 and not -12 because xyz=288 and not -288

x=12,y=6,z=4

sub the above values in 1/2x+1/4y+1/8z=m/n

and u will get the answer as m/n =11/96

since 11/96 is coprime m+n =107

Uahbid Dey
Aug 25, 2015

Is logarithms required?

The answer is 107.

5 solutions

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