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Quantitative Aptitude

Highest Power of a Number

Question 116 ...

Question

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

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

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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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