If logx/(b-c) = logy/(c-a) = logz/(a-b), then x^(b+c)y^(c+a)z^(a+b) = ?

Question

a) -1
b) 0
c) e
d) 2
e) 1

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Answer 1 · 2017-11-12T17:56:47

# 1.#

Supposing that the base of log is #m,# we have,

#log_mx/(b-c)=log_my/(c-a)=log_mz/(a-b)=k.#

#:. log_mx=k(b-c), log_my=k(c-a), log_mz=k(a-b).#

#:. x=m^(k(b-c)), y=m^(k(c-a)), z=m^(k(a-b)).#

#:. x^(b+c)={m^(k(b-c))}^(b+c)=m^{k(b-c)(b+c)}, i.e., #

# x^(b+c)=m^(k(b^2-c^2)).#

Similarly, #y^(c+a)=m^{k(c^2-a^2)} and z^(a+b)=m^{k(a^2-b^2)}.#

Therefore, #x^(b+c)*y^(c+a)*z^(a+b),#

#=m^(k(b^2-c^2))*m^{k(c^2-a^2)}*m^{k(a^2-b^2)},#

#=m^{k(b^2-c^2)+k(c^2-a^2)+k(a^2-b^2)},#

#=m^{k(b^2-c^2+c^2-a^2+a^2-b^2)},#

#=m^0,#

#=1.#

Enjoy Maths.!