Question

If a^p=bc, b^q=ac and c^r=ab then prove that pqr=p+q+r+2​

Answer 1

Hi,

I suppose
`a,b,c,p,q,r \ \in\ \mathbb{R}^+_0}`

Explanation:

`a^p=bc\Longrightarrow\ p\ ln(a)=ln(b)+ln(c)\\b^q=ac\Longrightarrow\ q\ ln(b)=ln(a)+ln(c)\\\Longrightarrow\ p\ ln(a)-ln(b)=q\ ln(b)-ln(a)\\\Longrightarrow\ (p+1)\ ln(a)=(q+1)\ ln(b)\\\\`

`c^r=ab\Longrightarrow\ r\ ln(c)=ln(a)+ln(b)\\r(p\ ln(a)-ln(b))=ln(a)+ln(b)\\ln(a)\ (rp-1)=ln(b)\ (r+1)\\\\\Longrightarrow\ (ln(a)\ (rp-1))/(ln(a)\ (p+1)) =(ln(b)\ (r+1))/(ln(b)\ (q+1))\\\\\Longrightarrow\ ((rp-1))/((p+1)) =((r+1))/((q+1))\\\\\Longrightarrow\ pqr+pr-q-1=pr+p+r+1\\\\\Longrightarrow\ pqr=p+q+r+2\\`