Question

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Answer 1

Question:

${9}^{x} \times {3}^{2} \times {3}^{ (\frac{ - x}{2} )^{ - 2} } = \frac{1}{27}$

To find:

• To find the value of x.

Solution:

${9}^{x} \times {3}^{2} \times {3}^{ (\frac{ - x}{2} )^{ - 2} } = \frac{1}{27} \\ = > ( {3}^{2})^{x} \times {3}^{2} \times {3}^{x} = \frac{1}{ {3}^{3} } \\ = > {3}^{2x} \times {3}^{2} \times {3}^{x} = {3}^{( - 3)} \\ = > {3}^{2x + 2 + x} = {3}^{( - 3)}$

• We know that when bases are same, powers are equal.

Therefore,

$2x \: + 2 \: + x \: = ( - 3) \\ \\ = > 3x \: + 2 = ( - 3) \\ = > 3x \: = ( - 3) - 2 \\ = > 3x \: = ( - 5) \\ = > x \: = \frac{( - 5)}{3}$

Answer:

• Therefore, value of x is (-5)/3

Formulas used:

1. ${a}^{m^{n} } = {a}^{mn}$
2. ${a}^{n} = {a}^{m^{n} } = {a}^{mn}$This is Only applicable when 'a' itself a perfect square, cube, etc.
3. ${a}^{m} \times {a}^{n} = {a}^{m + n}$
4. $\frac{1}{a ^{m} } = {a}^{ - m}$