Question

if 5^-p=4^-q=20^r;show that 1/p+1/q+1/r=0

Answer 1

$\textbf{ Hello Mate! }$

${5}^{ - p} = {4}^{ - q} = {20}^{r} = k \\ 5 = {k}^{ \frac{ - 1}{p} } \\ 4 = {k}^{ \frac{ - 1}{q} } \\ 20 = {k}^{ \frac{1}{r} } \\ \\ 4 \times 5 = 20 \\ {k}^{ \frac{ - 1}{p} } \times {k}^{ \frac{ - 1}{q} } = {k}^{ \frac{1}{r} } \\ \frac{ - 1}{p} + \frac{( - 1)}{q} = \frac{1}{r} \\ 0 = \frac{1}{r} + \frac{1}{p} + \frac{1}{q} \: or \\ \frac{1}{p} + \frac{1}{p} + \frac{1}{r} = 0$

$\boxed{ \textsf{ Hence\:proved }}$

Have great future ahead!

Answer 2

Question :-

$\sf \: \: if \: {5}^{ - p} = {4}^{ - q} = {20}^{r} \: \\ \sf \: prove \: that \: \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0$

Proof :-

We have ,

$\rightarrow \sf \: {5}^{ - p} = {4}^{ - q} = 20r \: = c(say)\\ \\ if \\ \implies \sf {5}^{ - p} = c \\ \sf taking \: \log \:on \: both \: sides\\ \implies \sf - p \log5 = \log c \\ \\ \implies \sf \log5 = - \frac{1}{p} \log c \: \: \: .....eq.(1) \\ if \: \\ \implies \sf {4}^{ - q} = c \\ \sf taking \: \log \: on \: both \: sides \\ \\ \implies \sf \log4 = \frac{ - 1}{q} \log c \: \: \: \: ....eq.(2)$

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now we have ,

$\rightarrow \sf \: {20}^{r} = c \\ \: \\ \sf \: taking \: \log \: on \: both \: sides \\ \\ \rightarrow \sf r \log20 = \log c \\ \\ \rightarrow \sf \log20 = \frac{1}{r} \log c \\ \\ \rightarrow \sf \log (5 \times 4) = \frac{1}{r} \log c \: \\ \sf \because \log \: mn = \log m + \log n \\ \therefore \\ \rightarrow \sf\log5 + \log4 = \frac{1}{r} \log c \\ \\ \sf from \: eq.(1) \: \: and \: eq.(2) \\ \\ \rightarrow \sf \frac{ - 1}{p} \log c - \frac{1}{q} \log c = \frac{1}{r} \log c \\ \\ \rightarrow \sf \log c \{ - \frac{1}{p} - \frac{1}{q} \} = \frac{1}{r} \log c \\ \\ \rightarrow \boxed{ \sf{ \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 0}}$

hence proved .

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