SOLUTION: If logx:3=logy:4=logz:5 then yz in terms of x is

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Question 1128680: If logx:3=logy:4=logz:5 then yz in terms of x is
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


log%28x%2C3%29=n --> x%5En=3
log%28y%2C4%29=n --> y%5En=4
log%28z%2C5%29=n --> z%5En=5

Then

%28%28y%5En%29%28z%5En%29%29%2F%28x%5En%29+=+20%2F3
%28y%5En%29%28z%5En%29+=+%2820%2F3%29%28x%5En%29
%28yz%29%5En+=+%2820%2F3%29%28x%5En%29
yz+=+x%2A%28%2820%2F3%29%5E%281%2Fn%29%29

Answer by ikleyn(52887) About Me  (Show Source):
You can put this solution on YOUR website!
.

            I read it in this way

                   log%28%28x%29%29%2F3 = log%28%28y%29%29%2F4 = log%28%28z%29%29%2F5.             (1)

            Also, the condition means that  x,  y  and  z  are positive real numbers  (by  DEFAULT).



Solution 

Multiply (1) by 60 = 3*4*5  (all the terms). You will get an equivalent equation


    20*log(x) = 15*log(y) = 12*log(z).


    x%5E20 = y%5E15 = z%5E12.


    From  x%5E20 = y%5E15  for real positive x and y  we obtain  y = x^(20/15) = x^(4/3).


    From  x%5E20 = z%5E12  for real positive x and z  we obtain  z = x^(20/12) = x^(5/3).


    Now  yz = x*(4/3)*x^(5/3) = x^(4/3+5/3) = x^(9/3) = x^3.


ANSWER.  "yz in terms of x"  is  yz = x^3.

Solved.