Question #f472a

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Answer 1 · 2015-06-09T19:54:36

#z= y/(3y/x+3/2)#

First, you want to turn the #x#, #y# and #z# powers into factors. To do so, use the naperian logarithm:

#color(gray)(Note: ln(a*b) = ln(a)+ln(b) and ln(a^b) = bln(a))#

#ln(2^x) = ln(3^y) = ln((24sqrt3)^z)#

#= xln(2) = yln(3) = zln(24sqrt3)#

Then, isolate #z#:

#z = (yln(3))/(ln(24sqrt3)) = (yln(3))/(ln(3*2^3)+ln(3^(1/2)))#

#= (yln(3))/(3ln(2)+ln(3)+ln(3)/2)#

Replace #ln(2)# with #y/xln(3)#

#z= (yln(3))/(3y/xln(3)+ln(3)+ln(3)/2)#

#= (ycancel(ln(3)))/((3y/x+1+1/2)cancel(ln(3))) = y/(3y/x+3/2)#