Question

How do you solve ${{81}^{x}}=9{{\left( 3 \right)}^{x}}$

Answer 1

Hint: We can solve the above-given question by using some simple transformations. By using the transformations, we can make the given question very easy to do. In this question, we have been asked to solve the given expression that is to find the value of the variable.

Complete step by step answer:
From the question given we have been asked to solve ${{81}^{x}}=9{{\left( 3 \right)}^{x}}$
As we have already discussed above, we have to do some simple transformations to get the question very easy to do.
From the question it had been given that ${{81}^{x}}=9{{\left( 3 \right)}^{x}}$
$\Rightarrow {{81}^{x}}=9{{\left( 3 \right)}^{x}}$
We know that $81$ can be written as ${{3}^{4}}$
By substituting it in the given question, we get the below equation ${{\left( {{3}^{4}} \right)}^{x}}=9{{\left( 3 \right)}^{x}}$
Now, we know the basic formula of exponents ${{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}$
By using the above basic formula of exponents, we get ${{3}^{4x}}=9{{\left( 3 \right)}^{x}}$
We know that $9$ can be written as ${{3}^{2}}$
By substituting it in the above equation, we get ${{3}^{4x}}={{3}^{2}}{{\left( 3 \right)}^{x}}$
Now, we also know another basic formula of exponents ${{a}^{x}}{{a}^{y}}={{a}^{x+y}}$
By using the above formula in the above equation, we get ${{3}^{4x}}={{3}^{2+x}}$
As the bases are equal in the above equation, we can equate the indices or powers.
By equating the indices or powers of the above equation, we get $4x=2+x$
On furthermore simplifying the above equation we get
$3x=2$
$\Rightarrow x=\dfrac{2}{3}$
Hence, the given equation is simplified.


Note:
 We should be well aware of the exponents and powers. We should be very careful while doing the calculation in this type of problem as the calculation part is somewhat difficult in this type of problem. We should know which type of transformation is to be used for the given question to get the question in an easier way. This can be simply done as ${{81}^{x}}=9{{\left( 3 \right)}^{x}}\Rightarrow {{9}^{2x}}={{9}^{1+\dfrac{x}{2}}}\Rightarrow 1+\dfrac{x}{2}=2x\Rightarrow x=\dfrac{2}{3}$