SOLUTION: If 2^x = 4^y = 8^z and 1/(2x) = 1/(4y) = 1/(8z) = 22/7. Show that x = 7/16, y = 7/32 and z = 7/48.

Algebra ->  Test -> SOLUTION: If 2^x = 4^y = 8^z and 1/(2x) = 1/(4y) = 1/(8z) = 22/7. Show that x = 7/16, y = 7/32 and z = 7/48.      Log On


   

Question 1140058: If 2^x = 4^y = 8^z and 1/(2x) = 1/(4y) = 1/(8z) = 22/7. Show that x = 7/16, y = 7/32 and z = 7/48.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given:
2%5Ex+=+4%5Ey+=+8%5Ez and
+1%2F%282x%29+=+1%2F%284y%29+=+1%2F%288z%29+=+22%2F7=>doesn't make sense, here must be the sum:+1%2F%282x%29%2B1%2F%284y%29+%2B+1%2F%288z%29+=+22%2F7
Show that:
x+=+7%2F16, y+=+7%2F32 and z+=+7%2F48

We know x%3C%3E0 , y%3C%3E0 and z%3C%3E0 because otherwise they would make the denominators zero, and the expression
1%2F%282x%29+%2B+1%2F%284y%29+%2B+1%2F%288z%29 would not exist as a real number.
As 4=2%5E2 and 8=2%5E3 , if 2%5Ex+=+4%5Ey+=+8%5Ez , 2%5Ex+=+4%5Ey+=+%282%5E2%29%5Ey+=+2%5E%282y%29 , and 2%5E%282y%29%2F2%5Ex=2%5E%282y-x%29=1+meaning+that+%7B%7B%7B2y-x=0 , x=2y , and y=%281%2F2%29x.
Also 2%5Ex+=+8%5Ez+=+%282%5E3%29%5Ez+=+2%5E%283z%29 , which means x=3z , and z=%281%2F3%29x .
You could also talk about taking logarithms base 2.\ to get to the same conclusions.
Then, substituting into 1%2F%282x%29+%2B+1%2F%284y%29+%2B+1%2F%288z%29+=+22%2F7 , we get
1%2F%282x%29+%2B+1%2F%284%281%2F2%29x%29+%2B+1%2F%288z%29+=+22%2F7
1%2F%282x%29+%2B+1%2F%282x%29+%2B+1%2F%288z%29+=+22%2F7
1%2Fx+%2B+1%2F%288z%29+=+22%2F7
1%2F%283z%29+%2B+1%2F%288z%29+=+22%2F7

Multiplying both sides of the equation times the non-zero number 3%2A8%2A7%2Az ,
we get the equivalent equation
8%2A7%2B3%2A7=22%2A3%2A8%2Az
%288%2B3%29%2A7=22%2A3%2A8%2Az
11%2A7=2%2A11%2A3%2A8%2Az
7%2F%282%2A3%2A8%29=z
highlight%28z=7%2F48%29
Then, as x=3z , x=3%2A7%2F%282%2A3%2A8%29=7%2F%282%2A8%29=highlight%287%2F16%29 ,
and as y=%281%2F2%29x , y=%281%2F2%29%2A%287%2F16%29=highlight%287%2F32%29