Question

Find the value of $m$ for which
(i) $5^m÷5^{-3}=5^5$

(ii) $4^m=64$

(iii) $8^{m-3}=1$

(iv) $(a^3{)}^m=a^9$

(v) $(5^m{)}^2\times (25{)}^3\times {125}^2=1$

(vi) $2^m=(8{)}^{\frac{1}{3}}÷(2^3{)}^{2/3}$

Answer 1

Pargt $\left(i\right)$

$5^m÷5^{-3}=5^5$

$5^m\times 5^3=5^5$

$5^m=5^{5-3}$

$5^m=5^2$

On comparing both sides, we get

$m=2$

Part $\left(ii\right)$

$4^m=64$

$4^m=4^3$

On comparing both sides, we get

$m=3$

Part $\left(iii\right)$

$8^{m-3}=1$

$\frac{8^m}{8^3}=1$

$8^m=8^3$

On comparing both sides, we get

$m=3$

Part $\left(iv\right)$

$(a^3{)}^m=a^9$

$a^{3m}=a^9$

On comparing both sides, we get

$3m=9$

$m=3$

Part $\left(v\right)$

$(5^m{)}^2\times (25{)}^3\times {125}^2=1$

$5^{2m}\times (5^2{)}^3\times {\left(5^3\right)}^2=1$

$5^{2m}\times \left(5^6\right)\times 5^6=1$

$5^{2m}\times 5^{12}=1$

$5^{2m}=5^{-12}$

On comparing both sides, we get

$2m=-12$

$m=-6$

Part $\left(vi\right)$

$2^m=(8{)}^{\frac{1}{3}}÷(2^3{)}^{2/3}$

$2^m=(2^3{)}^{\frac{1}{3}}÷2^2$

$2^m=2÷2^2$

$2^m=1÷2$

$2^m=2^{-1}$

On comparing both sides, we get

$m=-1$

Hence, this is the answer.