[tex] \sf{if \quad {a}^{p} = {b}^{q} = {c}^{r} = abc } \\ \bf{then} \: \: \: \green{\boxed{ \bf{ \orange{pqr}}} }= [/tex]
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[tex] \sf{if \quad {a}^{p} = {b}^{q} = {c}^{r} = abc } \\ \bf{then} \: \: \: \green{\boxed{ \bf{ \orange{pqr}}} }= [/tex]
quality answers needed hoping for great answers no copying from internet
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[tex] \huge{\red{\bf{answer}}}[/tex]
we have a^p = a^q = c^r = abc, so
[tex] {a}^{p} = abc \\ \\ a = (abc)\frac{1}{p} \: \: \: \: \: \: \: .1 \\ \\ \\ {a}^{q} = abc \\ \\ b = (abc)\frac{1}{q} \: \: \: \: \: .2 \\ \\ and \\ \\ {c}^{r} = abc \\ \\ {c}^{r} =(abc) \frac{1}{r}\: \: \: \: \: .3[/tex]
Now we multiple equation 1 , 2 and 3 and get;
[tex]abc = (abc) \frac{1}{p} + \frac{1}{q} + \frac{1}{c} [/tex]
( As we know (a^m) (a^n) = (a^m+n))
Be comparing both we side get;
[tex] \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1[/tex]
[tex] = \frac{qr + pr + pq}{pqr} = 1[/tex]
answer : pqr = qr + pr + pq
Verified answer
[tex]\underline{\underline{\bf{Given-}}}[/tex]
[tex] \sf{if \quad {a}^{p} = {b}^{q} = {c}^{r} = abc }[/tex]
[tex]\underline{\underline{\bf{To\:Find-}}}[/tex]
[tex] \sf{Value \: of \: PQR }[/tex]
[tex]\underline{\underline{\bf{Solution-}}}[/tex]
[tex] \sf{a^p = abc}[/tex]
[tex]\sf a = (abc)^ {\frac{1}{p}} \_ \: \_ \: \_ \: \_ \: \_ (1)[/tex]
[tex] \sf \quad \quad \quad \quad \quad \quad \: (since \: after \: transfering \: we \: will \: take \: the \: pth \: root )[/tex]
[tex] \tt{Similarly-}[/tex]
[tex] \sf{b^q = abc}[/tex]
[tex] \sf{b = (abc)^{\frac{1}{q}}\_ \: \_ \: \_ \: \_ \: \_ \: (2)}[/tex]
[tex] \tt{and}[/tex]
[tex] \sf{c^r = abc}[/tex]
[tex] \sf{c = (abc)^{\frac{1}{r}}\_ \: \_ \: \_ \: \_ \: \_ \: (3)}[/tex]
[tex] \underline{ \sf{Multiplying \: eq. (1),(2) and (3)}}[/tex]
[tex] \sf{abc = (abc)^{\frac{1}{p}}.(abc)^{\frac{1}{q}}.(abc)^{\frac{1}{r}}}[/tex]
[tex] \sf{abc = (abc)^{\frac{1}{p} +\frac{1}{q}+\frac{1}{r}}}[/tex]
[tex] \sf \quad \quad \quad \quad \quad \quad \: (a^m \times a^n = a^{m+n} )[/tex]
[tex] \underline{ \sf{Comparing \: powers \: of \: both \: sides-}}[/tex]
[tex] \sf1 = \dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}[/tex]
[tex] \sf{1 = \dfrac{qr+pr+pq}{pqr} }[/tex]
[tex]\boxed{\bf{pqr = qr+pr+pq}}[/tex]