Question

If \[{{\left( \dfrac{9}{7} \right)}^{3}}\times {{\left( \dfrac{49}{81} \right)}^{\left( 2x-6 \right)}}={{\left( \dfrac{7}{9} \right)}^{9}}\] then value of \[x\] is

Answer 1

Hint: In this type of question we have to use the concept of indices. We have to use different rules of indices such as \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\], \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\], \[{{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}\], \[{{x}^{-n}}=\dfrac{1}{{{x}^{n}}}\] etc. Also we know that if the bases are the same then we can compare the indices of the both sides of an equation. Here, we simplify the left side of the equation and then by comparing the indices of the both sides we can obtain the value of \[x\]

Complete step by step answer:
Now we have to find the value of \[x\] if \[{{\left( \dfrac{9}{7} \right)}^{3}}\times {{\left( \dfrac{49}{81} \right)}^{\left( 2x-6 \right)}}={{\left( \dfrac{7}{9} \right)}^{9}}\]
Let us consider the L.H.S. of the given equation
\[\Rightarrow L.H.S.={{\left( \dfrac{9}{7} \right)}^{3}}\times {{\left( \dfrac{49}{81} \right)}^{\left( 2x-6 \right)}}\]
Now, we know that, \[49\] can be expressed as \[{{7}^{2}}\] while \[81\] can be expressed as \[{{9}^{2}}\].
Hence, we can write
\[\Rightarrow L.H.S={{\left( \dfrac{9}{7} \right)}^{3}}\times {{\left( \dfrac{{{7}^{2}}}{{{9}^{2}}} \right)}^{\left( 2x-6 \right)}}\]
As we know that, \[{{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}\]
\[\Rightarrow L.H.S.=\dfrac{{{9}^{3}}}{{{7}^{3}}}\times \dfrac{{{\left( {{7}^{2}} \right)}^{\left( 2x-6 \right)}}}{{{\left( {{9}^{2}} \right)}^{\left( 2x-6 \right)}}}\]
Now, by using the rule \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\], we can write
\[\Rightarrow L.H.S.=\dfrac{{{9}^{3}}}{{{7}^{3}}}\times \dfrac{{{7}^{\left( 4x-12 \right)}}}{{{9}^{\left( 4x-12 \right)}}}\]
Now we will separate out the terms of \[9\] and \[7\] we get,
\[\Rightarrow L.H.S.=\dfrac{{{9}^{3}}}{{{9}^{\left( 4x-12 \right)}}}\times \dfrac{{{7}^{\left( 4x-12 \right)}}}{{{7}^{3}}}\]
By using the rule \[\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}\] we can simplify the above expression as
\[\Rightarrow L.H.S.={{9}^{3-\left( 4x-12 \right)}}\times {{7}^{4x-12-3}}\]
\[\begin{align}
  & \Rightarrow L.H.S.={{9}^{3-4x+12}}\times {{7}^{4x-15}} \\
 & \Rightarrow L.H.S.={{9}^{-4x+15}}\times {{7}^{4x-15}} \\
 & \Rightarrow L.H.S.={{9}^{-\left( 4x-15 \right)}}\times {{7}^{4x-15}} \\
\end{align}\]
As we know that, \[{{x}^{-n}}=\dfrac{1}{{{x}^{n}}}\] we can rewrite the above expression as
\[\begin{align}
  & \Rightarrow L.H.S.=\dfrac{1}{{{9}^{\left( 4x-15 \right)}}}\times {{7}^{\left( 4x-15 \right)}} \\
 & \Rightarrow L.H.S.=\dfrac{{{7}^{\left( 4x-15 \right)}}}{{{9}^{\left( 4x-15 \right)}}} \\
\end{align}\]
Now again by using the rule \[{{\left( \dfrac{a}{b} \right)}^{m}}=\dfrac{{{a}^{m}}}{{{b}^{m}}}\] we can write,
\[\Rightarrow L.H.S.={{\left( \dfrac{7}{9} \right)}^{\left( 4x-15 \right)}}\]
By given we have the R.H.S. of the equation as \[{{\left( \dfrac{7}{9} \right)}^{9}}\].
We know that, if the bases are the same then we can compare the indices of the both sides of an equation.
\[\Rightarrow \left( 4x-15 \right)=9\]
On simplification we get
\[\begin{align}
  & \Rightarrow 4x=9+15 \\
 & \Rightarrow 4x=24 \\
 & \Rightarrow x=\dfrac{24}{4} \\
 & \Rightarrow x=6 \\
\end{align}\]
Hence, the value of \[x\] if \[{{\left( \dfrac{9}{7} \right)}^{3}}\times {{\left( \dfrac{49}{81} \right)}^{\left( 2x-6 \right)}}={{\left( \dfrac{7}{9} \right)}^{9}}\] is \[6\].

Note: In this type of question students have to remember the different rules of indices. Also they have to be well familiar with the squares of the numbers from 1 to 10. Also students have to note that they have to compare the indices of R.H.S. and L.H.S. after simplifying the L.H.S.