Question

If x^1/a = y^1/b = z^1/c and xyz= 1 ,then prove that a+b+c=0 ( step by step)​

Answer 1

${\huge{\star}}\:\:\: {\bf{\LARGE{\underline{\bold{Given :}}}}}$

$\rightsquigarrow \sf x^\frac{1}{a}=y^\frac{1}{b}=z^\frac{1}{c}$

$\rightsquigarrow \sf xyz=1$

${\huge{\star}}\:\:\: {\bf{\LARGE{\underline{\bold{Proof :}}}}}$

$\circ\:\:\:\: \sf x^\frac{1}{a}=y^\frac{1}{b}=z^\frac{1}{c}=k\:\:\:\bf[let]$

$\begin{array}{ccc}\sf x^\frac{1}{a}=k&\sf y^\frac{1}{b}=k\:&\sf z^\frac{1}{c}=k\:\\ \sf\mapsto\log_xk=\frac{1}{a}&\mapsto \sf \log_yk=\frac{1}{b}&\sf\mapsto\log_zk=\frac{1}{c} \:\\ \mapsto \sf \log_kx=a&\mapsto\sf\log_ky=b&\mapsto\sf\log_kz=c&\underline{\:\:\:\:\:}(1)&\underline{\:\:\:\:\:}(2)&\underline{\:\:\:\:\:\:}(3) \end{array}$

now,

adding eqn (1) eqn (2) and eqn (3)

$\dashrightarrow \sf a+b+c = \log_kx+\log_ky+\log_kz$

$\dashrightarrow \sf (a+b+c)=\log_k(xyz)$

$\dashrightarrow \sf (a+b+c)=\log_k(1)$

$\dashrightarrow \underline{\boxed{\sf (a+b+c)=0}}$

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