Question

find the value of m for which 5^3m+2÷5^3=5^15​

Answer 1

Formula to be noted :

$\sf:\implies {a}^{x}\div{a}^{y} = {a}^{x -y}$

Solution :

$\longmapsto\sf{5}^{ \green{(3m + 2)}} \div {5}^{ \green3} ={5}^{ \green{15}}$

we know that :

$\sf:\implies {a}^{x}\div{a}^{y} = {a}^{x -y}$

therefore,

$\sf\longmapsto {5}^{3m +2- 3} = {5}^{15}$

$\sf\longmapsto {5}^{3m +2- 3 \: = \: 15}$

now lets solve the above exponents.

$\implies \sf3m +2- 3 =15$

$\implies \sf3m -1=15$

$\implies \sf3m=15 + 1$

$\implies \sf3m=16$

$\implies \sf \: m =\dfrac{16}{3}$

Answer 2

Given:

• ${\tt{ 5^{3m+2} ÷ 5^3 = 5^{15} }}$

To Find

• Value of the m?

Solution:

$\circ \: \: \: {\boxed{\tt\purple{a^m ÷ a^n = a^{m-n} }}}$

Or

$\circ \: \: \: {\boxed{\tt\red{ \dfrac{a^m}{a^n} = a^{m-n} }}}$

After substituting values,

$\\ \colon\leadsto{\tt{ 5^{3m+2} ÷ 5^3 = 5^{15} }} \\ \\ \\ \colon\leadsto{\tt{ \cancel{5} ^{3m + 2 - 3} = \cancel{5} ^{15} }} \\ \\ \\ \colon\leadsto{\tt{ 3m + 2 - 3 = 15 }} \\ \\ \\ \colon\leadsto{\tt{ 3m - 1 = 15 }} \\ \\ \\ \colon\leadsto{\tt{ 3m = 15 + 1 }} \\ \\ \\ \colon\leadsto{\tt{ 3m = 16 }} \\ \\ \\ \colon\leadsto{\tt{ m = \cancel{ \dfrac{16}{3} } }} \\ \\ \\ \colon\leadsto{\boxed{\mathfrak\pink{ m = 5.3 }}}$

Hence,

• The Value of m is 5.3

More to Know:

• $\circ \: {\boxed{\tt\red{ \dfrac{a^m}{a^n} = a^{m-n} }}}$

• $\circ \: {\boxed{\tt\red{ a^m \times a^n = a^{m+n} }}}$

• $\circ \: {\boxed{\tt\red{ (a^m)^n = a^{mn} }}}$

• $\circ \: {\boxed{\tt\red{ (ab)^n = a^nb^n }}}$

• $\circ \: {\boxed{\tt\red{ a^{mn} \neq (a^m)^n }}}$