Question

If ${9^x} \times {3^2} \times {\left( {{3^{\dfrac{{ - x}}{2}}}} \right)^{ - 2}} = \dfrac{1}{{27}}$, find x.

Answer 1

Hint: In the left hand side we have 3 terms, two terms with base 3 and one term with base 9. Convert the term with base 9 to the term with base 3. So the left hand side will now have all powers with base 3. Convert the right hand side value also to a power with base 3. Now, the bases are equal in both LHS and RHS now. So the exponents of the left hand side and right hand side must be equal. Equate the exponents and find the value of x.

Complete step-by-step answer:
We are given an expression ${9^x} \times {3^2} \times {\left( {{3^{\dfrac{{ - x}}{2}}}} \right)^{ - 2}} = \dfrac{1}{{27}}$ and we have to find the value of x.
In the LHS of ${9^x} \times {3^2} \times {\left( {{3^{\dfrac{{ - x}}{2}}}} \right)^{ - 2}} = \dfrac{1}{{27}}$, we have 3 terms with base 9, base 3 and base 3.
Convert the term with base 9 to the terms with base 3.
$
  9 = {3^2} \\
  {9^x} = {\left( {{3^2}} \right)^x} = {3^{2x}} \\
  \left( {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right) \\
  {\left( {{3^{\dfrac{{ - x}}{2}}}} \right)^{ - 2}} = {3^{\dfrac{{ - x}}{2} \times - 2}} = {3^x} \\
  \left( {\because {{\left( {{a^m}} \right)}^n} = {a^{mn}}} \right) \\
  {9^x} \times {3^2} \times {\left( {{3^{\dfrac{{ - x}}{2}}}} \right)^{ - 2}} = {3^{2x}} \times {3^2} \times {3^x} \\
 $
Now the LHS of ${9^x} \times {3^2} \times {\left( {{3^{\dfrac{{ - x}}{2}}}} \right)^{ - 2}} = \dfrac{1}{{27}}$ have all the terms in base 3, ${3^{2x}} \times {3^2} \times {3^x}$
Converting the RHS value into a power with base 3
$
  27 = 3 \times 3 \times 3 = {3^3} \\
  \dfrac{1}{{27}} = \dfrac{1}{{{3^3}}} = {3^{ - 3}} \\
  \left( {\because \dfrac{1}{{{a^m}}} = {a^{ - m}}} \right) \\
 $
Now in both the LHS and RHS, the base of the powers is 3.
$
  {3^{2x}} \times {3^2} \times {3^x} = {3^{ - 3}} \\
   \to {3^{2x + 2 + x}} = {3^{ - 3}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
 $
Hence the bases are equal in LHS and RHS. So the exponents must be equal.
$
   \to 2x + 2 + x = - 3 \\
   \to 3x = - 3 - 2 \\
   \to 3x = - 5 \\
  \therefore x = \dfrac{{ - 5}}{3} \\
 $
Therefore, the value of x is $\dfrac{{ - 5}}{3}$

Note: A positive exponent means repeated multiplication of the base with the base, a negative exponent means repeated division of the base by the base. When the negative exponents are moved to the denominator, the exponent becomes positive and vice-versa. Be careful with the positive and negative exponents as the negative exponents are always the reciprocals of positive exponents. And if you are unsure the result you got, just verify your answer by substituting it in the given expression. In the above solution, to verify the answer you can substitute the value of x in the given equation.