Question

Find the value of m for which
$3 ^{2m + 1} \div 9 = 27$
Pls help urgent
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Answer 1

GIVEN :-

$\\ \implies \sf \: (3)^{2m + 1} \div 9 = 27$

TO FIND :-

• The value of m.

SOLUTION :-

$\\ \implies \sf \: (3)^{2m + 1} \div 9 = 27$

➔ 9 can be written as (3)².

$\\ \implies \sf \: (3) ^{2m + 1} \div (3) ^{2} = 27$

➔ Using identity:- $\sf a^{m} \div a^{n} = a^{m-n}$

$\\ \implies \sf \: ( 3 ) ^{2m + 1 - 2} = 27 \\ \\ \implies\sf \: (3) ^{2m - 1} = 27$

➔ 27 can be written as (3)³.

$\\ \implies \sf \: (3) ^{2m - 1} = (3) ^{3}$

➔ Using Identity:- $\sf a^{b} = a^{c} \to b = c$

$\\ \implies \sf \: 2m - 1 = 3 \\ \\ \implies \sf \: 2m = 3 + 1 \\ \\ \implies \sf \: 2m = 4 \\ \\ \implies \sf \: m = \dfrac{4}{2} \\ \\\implies\underline{ \boxed{ \red{ \sf \:m = 2 }}}$

❏ Hence the value of m is 2.

ADDITIONAL INFORMATION :-

$\boxed{\begin{minipage}{7cm} \\ \sf{ \implies \bf \sqrt[n]{ \sqrt[m]{ \sqrt[p]{((a^{x} )^{y}) ^{z} } } } = (a ^{xyz} )^{ \frac{1}{mnp} } = a ^{ \frac{xyz}{mnp} } } \\ \\ \sf{ \implies a^m \times a^n = a^{m+n} } \\ \\ \sf{ \implies {a}^{m} \times b^m = ab^m } \\ \\ \sf{ \implies \dfrac{a^m}{a^n} = a^{m - n} ( \tt{ If \: m > n} ) } \\ \\ \sf{ \implies \dfrac{a^m}{ a^n} = \dfrac{ 1}{ a^{n-m} } ( \tt{ If \: n > m )} } \\ \\ \sf{ \implies (a^m)^n = a^{mn} } \\ \\ \sf{ \implies a^{-n} = \dfrac{1}{ a^n} }\end{minipage}}$

Answer 2

Answer:

${3}^{2m + 1} \div 9 = 27 \\ \\ = > {3}^{2m + 1} \div {3}^{2} = {3}^{3} \\ \\ = > {3}^{2m - 1} = {3}^{3} \\ \\ = > 2m - 1 = 3 \\ \\ = > 2m = 4 \\ \\ = > m = \frac{4}{2} = 2$