Question

If a, b, c are positive real numbers other than unity such that :
$\frac{a(b+c-a)}{loga}$ = $\frac{b(c+a-b)}{logb}$ = $\frac{c(a+b-c)}{logc}$
Prove that a${}^b$ b${}^a$ = b${}^c$ c${}^b$ = c${}^a$ a${}^c$.

Answer 1

$Let,\frac{a\left(b+c-a\right)}{\log a}=\frac{b\left(c+a-b\right)}{logb}=\frac{c\left(a+b-a\right)}{logc}=\frac{1}{x}$

$Atfirstwetake\frac{a\left(b+c-a\right)}{\log a}=k\log a=ka\left(b+c-a\right)$

$Onmultiplyingbothsidesbybb\log a=kab\left(b+c-a\right)---\left(1\right)and,\frac{b\left(a+c-b\right)}{logb}=\frac{1}{k}\log b-kb\left(c+a-b\right)$

$Onmultiplingbyabothsidesa\log b=kac+k{a}^{2}b-ka{b}^2---\left(2\right)$

$Now,adequation1and2b\log a+a\log b=2kabc\left(asweknowthat\log a^b=b\log a\right)\log \left(a^b\times b^a\right)=2kabc---\left(3\right)$

$Similarly\frac{b\left(a+c-b\right)}{\log b}=\frac{1}{k}\log b=kb\left(c+a-b\right)$

$Onmultiplingbycbothsidec\log b=kcb\left(c+a-b\right)c\log b=kb{c}^2+kabc-ke{b}^2---\left(4\right)and,\frac{c\left(a+b-c\right)}{\log c}=\frac{1}{k}\log c=kc\left(a+b-c\right)$

$Onmultiplyingbybbothsideboc=kabc\left(a+b-c\right)b\log c=kabc+k{b}^{2}c-kb{c}^2--\left(5\right)$

$Now,addeq{u}^{n}4and5clogb+blogc=2kabclog{b}^c+\log c^b=2kabc\log \left(b^c\times c^b\right)=2kabc----\left(6\right)$

$Samewayweget\log \left(c^a\times a^c\right)=2kabc----\left(7\right)\log \left(a^b\times b^a\right)=\log \left(b^c\times c^b\right)=\log \left(c^a\times a^c\right)$

$Ontakinganti-\log a^b\times b^a=b^c\times c^b=c^a\times a^{c}proved.$