Question

If a^p = b^q = c^r and b square = ac,then show that q=2rp by r+p

Answer 1

$Answer \: \\ \\ a {}^{p} = b {}^{q} = c {}^{r} \: = k \: \: \: \: \: \: let \\ \\ a {}^{p} = k \: \: \: \: \: \: \: \: \: b {}^{q} = k \: \: \: \: \: \: \: \: c {}^{r} = k \: \: \: \: \: \: \\ \\ a = k {}^{ \frac{1}{p} } \: \: \: \: ...Equation \: \: \: \: i \\ \\ b = k {}^{ \frac{1}{q} } \: \: ... \: Equation \: \: \: ii \\ \\ c = k {}^{ \frac{1}{r} } \: \: \: ... \: Equation \: \: \: iii \\ \\ \\ b {}^{2} = ac \: \: \: \: \: put \: the \: value \: of \: \: a \: \: \: b \: \: \: and \: \: \: \: c \: \\ in \: this \: given \: \: equation \: we \: have \\ \\ k {}^{ (\frac{2}{q}) } = k {}^{ \frac{1}{p} } \: \: \times \: \: k {}^{ \frac{1}{r} } \\ \\ k {}^{( \frac{2}{q} ) } = k {}^{ (\frac{1}{p} + \frac{1}{r}) } \\ \\ Now \: \: compare \: powers \: of \: k \: we \: have \\ \\ \\ \frac{2}{q} = \frac{1}{p} + \frac{1}{r} \\ \\ \frac{2}{q} = \frac{(r + p)}{rp} \\ \\ 2rp = q(r + p) \\ \\ q = \frac{2rp}{(r + p)} \\ \\ therefore \: \: \: q \: = \frac{2rp}{(r + p)}$