Solveeit Logo

Question

Mathematics Question on Laws of Exponents

Simplify and express the result in power notation with positive exponent.

  1. (−4) 5 ÷ (−4) 8
  2. (123)\left(\frac{1 }{ 2^3}\right) 2
  3. (−3) 4 × (53\frac{5}{3}) 4
  4. (3 -7 ÷ 3 -10) × 3 -5
  5. 2 -3 × (−7) -3
Answer

(i) (−4)5 ÷ (−4)8
Since aman\frac{a^m}{a^n}= am − n
(−4)5 ÷ (−4)8 = (4)5(4)8\frac{(−4)^5}{(−4)^8}= (−4)5−8
(−4)−3
=1(4)3= \frac{1}{(−4)^3}


(ii) (123)2\left(\frac{1}{2^3}\right)^2
Since, (am)n = amn
(123)2⇒ \left(\frac{1}{2^3}\right)^2
=126= \frac{1}{2^6}


(iii) (−3)4 ×(53)4× \left(\frac{5}{3}\right)^4
We know that am × bm =(ab)m and (ab\frac{a}{b})m = ambm\frac{a^m}{b^m} where a & b are non-zero integers and m is an integer
(−3)4 ×(53)4× \left(\frac{5}{3}\right)^4

(1×3)4×5434⇒ (−1 × 3)^4 × \frac{5^4}{3^4}

(−1)4 × 54
= 54 [∵(−1)4 = 1]


(iv)(3−7 ÷ 3−10) × 3−5
We know. aman=\frac{a^m}{a^n} = am − n and am × an = am + n
(3−7 ÷ 3−10) × 3−5 = (3−7−(−10)) × 3−5
= (3−7 + 10) × 3−5 = 33 × (3−5)
= 33 + (−5)
= 3−2 =132 \frac{1}{3^2}


(v) 2−3 × (−7)−3
We know that, am × bm = (ab)m
2−3 × (−7)−3
= [2 × (−7)]−3
= (−14)−3
=1(14)3= \frac{1}{(-14)^3} [Since a−m = 1/am]