Chapter 8: Exponents and Powers

Ans: False

LHS = (- 5)-2 x (-5)-3

Using law of exponents, am × an = am+n

Therefore, (–5)–2 × (–5)–3 = (–5)–5

Ans: False

LHS = (-4)-4 x (4)-1

Using law of exponents, am × an = am+n

(-4)-4 × (4)-1 = (4)-5

Ans: False

${\text{LHS  = }}{\left( {\dfrac{2}{3}} \right)^{ - 2}} \times {\left( {\dfrac{2}{3}} \right)^{ - 5}} = {\left( {\dfrac{2}{3}} \right)^{ - 2 - 5}} = {\left( {\dfrac{2}{3}} \right)^{ - 7}}$

So, L H S ≠ R H S

Ans: False.

Using law of exponent = a0 = 1

50 = 1.

Ans: False.

Using law of exponent = a0 = 1

(-2)0 = 1.

Ans: False.

Using law of exponent = a0 = 1

${\left( { - \dfrac{8}{2}} \right)^0} = 1$

Ans: False.

Using law of exponent = a0 = 1

(-6)0 = 1.

Ans: True.

LHS = (-7)4 x (-7)2

Using law of exponents, am × an = am+n

Therefore, (-7)-4 x (-7)-2 = (-7)-4+2 = (-7)-2

LHS = RHS

Ans: True

Using law of exponents, ${x^{ - m}} = \dfrac{1}{{{x^m}}}$

Thus, $\dfrac{1}{{{4^{ - 2}}}} = {4^2} = 4 \times 4 = 16$

Ans: False

Using law of exponents, ${x^{ - m}} = \dfrac{1}{{{x^m}}}$

Thus,${4^{ - 3}} = \dfrac{1}{{{4^3}}} = \dfrac{1}{{{{\left( {{2^2}} \right)}^3}}} = \dfrac{1}{{{2^6}}}$