Question 69 Find m- so that 2-93-2-96-2-92 m -1

We have (2/9 )^3 ×(2/9 )^6 = 2/9^2m 1 ⇒(2/9 )^3+6= (2/9 )^2m 1 ⇒(2/9 )^9 = (2/9 )^2m 1 ⇒ 9 = 2m 1 (since bases are equal, exponents can be equated) ⇒ 9+1 = ...

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Standard VIII

Mathematics

Exponents with like Bases

Question 69 F...

Question

Question 69

Find $m$ , so that ${(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{6} = {(\frac{2}{9})}^{2 m - 1}$

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Solution

We have ${(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{6} = {\frac{2}{9}}^{2 m - 1} \Rightarrow {(\frac{2}{9})}^{3 + 6} = {(\frac{2}{9})}^{2 m - 1} \Rightarrow {(\frac{2}{9})}^{9} = {(\frac{2}{9})}^{2 m - 1} \Rightarrow 9 = 2 m - 1 (since bases are equal, exponents can be equated) \Rightarrow 9 + 1 = 2 m \Rightarrow 10 = 2 m \Rightarrow \frac{10}{2} = \frac{2 m}{2} \Rightarrow 5 = m$

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Similar questions

Q. Find the value of

m

{(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{6} = {(\frac{2}{9})}^{2 m - 1}

Q. Find the value of

m

if :

{(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{- 6} = {(\frac{2}{9})}^{2 m - 1}

Question 116

Find x, so that

${(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{- 6} = {(\frac{2}{9})}^{2 x - 1} .$

Question 116

Find x, so that

${(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{- 6} = {(\frac{2}{9})}^{2 x - 1} .$

Question 2 (i)
Find: $9^{\frac{3}{2}}$

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Exponents with like Bases

Standard VIII Mathematics

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Question 69 Find m, so that (29)3×(29)6=(29)2m−1

We have (29)3×(29)6=292m−1⇒(29)3+6=(29)2m−1⇒(29)9=(29)2m−1⇒9=2m−1 (since bases are equal, exponents can be equated)⇒9+1=2m⇒10=2m⇒102=2m2⇒5=m

Question 69

Find $m$ , so that ${(\frac{2}{9})}^{3} \times {(\frac{2}{9})}^{6} = {(\frac{2}{9})}^{2 m - 1}$