Alg9.4.48

If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}\,$ prove that $a^{b+c}\cdot b^{c+a}\cdot c^{a+b}=1\,$

Let

$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k\,$

$\log a=k(b-c),\log b=k(c-a),\log c=k(a-b)\,$

To prove that

$a^{b+c}\cdot b^{c+a}\cdot c^{a+b}=1\,$

Applying log on both sides,

$(b+c)log a+(c+a)log b+(a+b)\log c=log1=0\,$

Now substituting the values of $\log a,\log b,\log c\,$ in the above

$k(b+c)(b-c)+k(c+a)(c-a)+k(a+b)(a-b)\,$

Simplifying

$k(b^2-c^2+c^2+a^2+a^2-b^2)\,$

Hence the result is zero in the above,which is proved.