Question

if 5 raise to the power minus p is equal to 4 raise to the power minus Q is equal to 20 raised to the power r show that one upon P + 1 upon Q + 1 upon R is equal to zero<br />​

Answer 1

$\textbf{Given:}$

$5^{-p}=4^{-q}=20^r$

$\textbf{To find:}$

$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}=0$

$\textbf{Solution:}$

$\text{Consider,}$

$5^{-p}=4^{-q}=20^r=k\;(\text{say})$

$5^{-p}=k\,\implies\,5=k^{\frac{-1}{p}}$

$4^{-q}=k\,\implies\,4=k^{\frac{-1}{q}}$

$\text{Now,\; 20^r=k }$

$(5{\times}4)^r=k$

$(k^{\frac{-1}{p}}{\times}k^{\frac{-1}{q}})^r=k$

$k^{(-{\frac{1}{p}-\frac{1}{q})}r}=k$

$k^{-\frac{r}{p}-\frac{r}{q}}=k$

$\text{Equating powers on bothsides we get}$

$-\dfrac{r}{p}-\dfrac{r}{q}=1$

$-\dfrac{1}{p}-\dfrac{1}{q}=\dfrac{1}{r}$

$\implies\bf\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}=0$

$\therefore\textbf{Hence proved}$

Find more:

2 a is equal to 3 b is equal to 4 c then a ratio b ratio c is

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Answer 2

Given:

5

−

�

=

4

−

�

=

2

0

�

5

−p

=4

−q

=20

r

To

find:

To find:

1

�

+

1

�

+

1

�

=

0

p

1

+

q

1

+

r

1

=0

Solution:

Solution:

Consider,

Consider,

5

−

�

=

4

−

�

=

2

0

�

=

�

(

say

)

5

−p

=4

−q

=20

r

=k(say)

5

−

�

=

�

⟹

5

=

�

−

1

�

5

−p

=k⟹5=k

p

−1

4

−

�

=

�

⟹

4

=

�

−

1

�

4

−q

=k⟹4=k

q

−1

\text{Now,\;\$20^r=k\$}

(

5

×

4

)

�

=

�

(5×4)

r

=k

(

�

−

1

�

×

�

−

1

�

)

�

=

�

(k

p

−1

×k

q

−1

)

r

=k

�

(

−

1

�

−

1

�

)

�

=

�

k

(−

p

1

−

q

1

)r

=k

�

−

�

�

−

�

�

=

�

k

−

p

r

−

q

r

=k

Equating powers on bothsides we get

Equating powers on bothsides we get

−

�

�

−

�

�

=

1

−

p

r

−

q

r

=1

−

1

�

−

1

�

=

1

�

−

p

1

−

q

1

=

r

1

⟹

1

�

+

1

�

+

1

�

=

0

⟹

p

1

+

q

1

+

r

1

=0

∴

Hence

proved

∴Hence proved

Find more:

2 a is equal to 3 b is equal to 4 c then a ratio b ratio c is