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if a^x=b , b^y=c , c^z=a show that xyz=1( abc positive numbers)
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a ^xyz = a so xyz = 1
x=logb/loga y=logc/logb z=loga/logc
xyz= logb/loga x logc/logb x loga/logc =1
how a^xyz=1
x/(b+c-a) = z/(a+b-c) => (a+b-c)x = (b+c-a)z x/(b+c-a) = y/(c+a-b) => (c+a-b)x = (b+c-a)y So (b+c-a)(z-y) = 2(b-c)x
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a^x=b and b^y = c therefore a^xy = c a^xy = c and c^z = a therefore a^xyz = a now a^1 = a therefore xyz = 1
thamkz a lot.i got it..
good
2^2^x=16^263x solve
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