Question

If 5^-p=4^-q=20r . Prove that 1/p+1/q+1/r=0​

Answer 1

Question:

If 5^(-p) = 4^(-q) = 20^r , then prove that ;

1/p + 1/q + 1/r = 0.

Note:

✓ log(A•B) = log(A) + log(B)

✓ log(A/B) = log(A) - log(B)

✓ log(A^B) = B•log(A)

Solution:

Let ,

5^(-p) = 4^(-q) = 20^r = k (say) -------(1)

{where k is any constant}

Now,

=> 5^(-p) = k {using eq-(1)}

=> log{5^(-p)} = log(k)

=> (-p)•log(5) = log(k)

=> log(5) = (-1/p)•log(k) --------(2)

Similarly,

=> 4^(-q) = k {using eq-(1)}

=> log{4^(-q)} = log(k)

=> (-q)•log(4) = log(k)

=> log(4) = (-1/q)•log(k) ----------(3)

Similarly,

=> 20^r = k {using eq-(1)}

=> log{20^r} = log(k)

=> r•log(20) = log(k)

=> log(20) = (1/r)•log(k)

=> log(5•4) = (1/r)•log(k)

=> log(5) + log(4) = (1/r)•log(k)

=> (-1/p)•log(k) + (-1/q)•log(k) = (1/r)•log(k)

{using eq-(2) and eq-(3)}

=> (-1/p -1/q)•log(k) = (1/r)•log(k)

=> -1/p -1/q = 1/r

=> 0 = 1/p + 1/q + 1/r

=> 1/p + 1/q + 1/r = 0

Hence proved.

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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