[(4)^ 1 (5)^ 1P× (5/8)^ 1] = [1/4 P/5×8/5] P = 1/4 = 8P/25 8P/25 = 1/4 32 P = 25 ∴ P = 25/32 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Mathematics

Power of a Power

[4-1-5-1P× 5/...

Question

$[(4)^{- 1} - (5)^{- 1} P \times (\frac{5}{8})^{- 1}]$

Open in App

Solution

$[(4)^{- 1} - (5)^{- 1} P \times (\frac{5}{8})^{- 1}]$
$= [\frac{1}{4} - \frac{P}{5} \times \frac{8}{5}]$
$P = \frac{1}{4} = \frac{8 P}{25}$
$\frac{8 P}{25} = \frac{1}{4}$
$32 P = 25$
$∴ P = \frac{25}{32}$

Suggest Corrections

Similar questions

Question 2

Complete the entries of the third column of the table:

$\begin{matrix} S.No & Equation & Value of Variable & Equation satisfied Yes/No (a) & 10 y = 80 & y = 10 (b) & 10 y = 80 & y = 8 (c) & 10 y = 80 & y = 5 (d) & 4 l = 20 & l = 20 (e) & 4 l = 20 & l = 80 (f) & 4 l = 20 & l = 5 (g) & b + 5 = 9 & b = 5 (h) & b + 5 = 9 & b = 9 (i) & b + 5 = 9 & b = 4 (j) & h - 8 = 5 & h = 13 (k) & h - 8 = 5 & h = 8 (l) & h - 8 = 5 & h = 0 (m) & p + 3 = 1 & p = 3 (n) & p + 3 = 1 & p = 1 (o) & p + 3 = 1 & p = 0 (p) & p + 3 = 1 & p = - 1 (q) & p + 3 = 1 & p = - 2 \end{matrix}$

Q. Evaluate:
(i)

{(\frac{1}{3})}^{- 1} - {(\frac{1}{4})}^{- 1}

(ii)

{(\frac{5}{8})}^{- 7} \times {(\frac{8}{5})}^{- 4}

Find :
1) $\frac{2}{5} \times \frac{1}{8}$

ii) $\frac{1}{5} \times \frac{3}{4}$

iii) $\frac{5}{8} \times \frac{8}{15}$

Evaluate (i) ${(\frac{1}{3})^{- 1} - (\frac{1}{4})^{- 1}}^{- 1}$ (ii) $(\frac{5}{8})^{- 7} \times (\frac{8}{5})^{- 4}$

Q. Simplify:

{[{(4)}^{- 1} - {(5)}^{- 1}]}^{2} \times {(\frac{5}{8})}^{- 1}

Join BYJU'S Learning Program

[(4)−1−(5)−1P×(58)−1]

[(4)−1−(5)−1P×(58)−1]=[14−P5×85]P=14=8P258P25=1432P=25∴P=2532

$[(4)^{- 1} - (5)^{- 1} P \times (\frac{5}{8})^{- 1}]$

$[(4)^{- 1} - (5)^{- 1} P \times (\frac{5}{8})^{- 1}]$
$= [\frac{1}{4} - \frac{P}{5} \times \frac{8}{5}]$
$P = \frac{1}{4} = \frac{8 P}{25}$
$\frac{8 P}{25} = \frac{1}{4}$
$32 P = 25$
$∴ P = \frac{25}{32}$