# Simplify.

(i) (25×t^{−4})/(5^{−3}×10×t^{−8}) (t ≠ 0)

(ii) (3^{−5}×10^{−5}×125)/(5^{−7}×6^{−5})

**Solution:**

The exponent of a number shows how many times the number is multiplied by itself.

(i) (25 × t^{−4})/(5^{−3 }× 10 × t^{−8})

Let's express 25 and 10 in terms of their prime actors as shown below,

= (5^{2 }× t^{−4})/(5^{−3 }× 5 × 2 × t^{−8 })

= (5^{2 }× t^{−4})/(5^{−3 + 1 }× 2 × t^{−8}) [Since, a^{m }× a^{n }= a^{m + n}]

= (5^{2 }× t^{−4})/(5^{−2 }× 2 × t^{−8})

= (5^{2−(−2) }× t^{−4−(−8)})/2 [Since, a^{m}/a^{n }= a^{m − n}]

= (5^{4 }× t^{−4 + 8})/2

= 625t^{4}/2

(ii) (3^{−5}×10^{−5}×125)/(5^{−7}× 6^{−5})

Let's express 10, 6 and 125 in terms of their prime factors.

= (3^{−5 }× (2 × 5)^{−5 }× 5^{3})/(5^{−7}× (2 × 3)^{-5})

= 3^{−5−(−5) }× 2^{−5−(−5) }× 5^{−5−(−7)+3} [Since, a^{m }× a^{n }= a^{m + n} and a^{m}/a^{n }= a^{m − n}]

= 3^{0 }× 2^{0 }× 5^{5 }

= 1 × 1 × 5^{5 }[∵a^{0}=1]

= 5^{5}

**Video Solution:**

## Simplify. (i) (25×t⁻⁴)/(5⁻³×10×t⁻⁸) (t ≠ 0) (ii) (3⁻⁵×10⁻⁵×125)/(5⁻⁷×6⁻⁵)

### Class 8 Maths NCERT Solutions - Chapter 12 Exercise 12.1 Question 7

**Summary:**

The value of the following expressions (i) (25×t^{−4})/(5^{−3}×10×t^{−8}) (t ≠ 0) (ii) (3^{−5}×10^{−5}×125)/(5^{−7}×6^{−5}) are (i) 625t^{4}/2 and (ii) 5^{5} respectively