Question

if log a /b-c =log b/c-a=log c /a-b, then find the value of abc . also prove that a^a. b^b.c^c=1,(a,b>0)
please answer this question it's urgent.
I will mark as brainliest.

Answer 1

logab−c=logbc−a=logca−b=klog⁡ab−c=log⁡bc−a=log⁡ca−b=k

logab−c=k⟹a=e(b−c)klog⁡ab−c=k⟹a=e(b−c)k

logbc−a=k⟹b=e(c−a)klog⁡bc−a=k⟹b=e(c−a)k

logca−b=k⟹c=e(a−b)klog⁡ca−b=k⟹c=e(a−b)k

Hence,

aabbcc=(e(b−c)k)a(e(c−a)k)b(e(a−b)k)caabbcc=(e(b−c)k)a(e(c−a)k)b(e(a−b)k)c

=(ea(b−c)k)(eb(c−a)k)(ec(a−b)k)=(ea(b−c)k)(eb(c−a)k)(ec(a−b)k)

=ea(b−c)k+b(c−a)k+c(a−b)k=ea(b−c)k+b(c−a)k+c(a−b)k

=e(ab−ac+bc−ab+ac−bc)k=e(ab−ac+bc−ab+ac−bc)k

=e0=e0

=1=1


hope it help mark as brainliest

Answer 2

Hlo mate your solution