Question 116 Find x- so that2-93-2-9-6-2-92 x -1

Given, (2/9 )^3 ×(2/9 )^ 6 = (2/9 )^2x 1. Using the law of exponents, a^m × a^n = (a)^m+n (2/9 )^3 6 = (2/9 )^2x 1 ⇒(2/9 )^ 3 = (2/9 )^2x 1 Since, the base ...

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Question 116

Find x, so that

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

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Solution

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

Using the law of exponents,
$a^{m} \times a^{n} = (a)^{m + n}$

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

Since, the base is same on both sides, we can equate the exponents.
$\Rightarrow - 3 = 2 x - 1 \Rightarrow 2 x = - 3 + 1 \Rightarrow 2 x = - 2 \Rightarrow x = - 1$

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Question 116

Find x, so that

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

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

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$\frac{1}{6}$ and $\frac{1}{9}$

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Question 116 Find x, so that (29)3×(29)−6=(29)2x−1.

Given, (29)3×(29)−6=(29)2x−1. Using the law of exponents, am×an=(a)m+n (29)3−6=(29)2x−1⇒(29)−3=(29)2x−1 Since, the base is same on both sides, we can equate the exponents. ⇒−3=2x−1⇒2x=−3+1⇒2x=−2⇒x=−1

Question 116

Find x, so that

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