Question

Find the value of $x$ , if $9\times {{3}^{x}}={{27}^{2x-3}}$

Answer 1

Hint: For finding the value of $x$ in the given question where $x$ is used in power of a number, first of all we need to make the base of $x$ same on both sides of equal sign so that we can make these powers equal by comparison. After comparison we will get a polynomial equation. So, we need to solve this equation to find the value of $x$ .

Complete step-by-step solution:
Here is the given equation where we need to find the value of $x$ as:
$\Rightarrow 9\times {{3}^{x}}={{27}^{2x-3}}$
Since, we can see that both $9$ and $27$ are multiple of $3$ . So, we can write them in the form of squares and cubes for $3$ . Here, $9$ is equal to ${{3}^{2}}$ and $27$ is equal to ${{3}^{3}}$ . Thus, we can write the above equation as:
$\Rightarrow {{3}^{2}}\times {{3}^{x}}={{\left( {{3}^{3}} \right)}^{2x-3}}$
If the base is the same and power differs, we follow some rules related to the base and the power. Here, according to the above step, the first rule is the product of same base with different power can be written with one base with sum of powers like ${{x}^{a}}\times {{x}^{b}}={{x}^{a+b}}$ and if the power of any base has a power, we can write those powers in multiplication form as ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{a\times b}}$ . So, we will use these rules in the above step to go further in the process of solution as:
$\Rightarrow {{3}^{2+x}}={{3}^{3\times \left( 2x-3 \right)}}$
Now, we open the bracket of power of the term of the right hand side as:
$\Rightarrow {{3}^{2+x}}={{3}^{3\times 2x-3\times 3}}$
After simplification, we will get in power:
$\Rightarrow {{3}^{2+x}}={{3}^{6x-9}}$
Since, base is the same for both sides of the equation. So, we can write the above equation as:
$\Rightarrow 2+x=6x-9$
Now, we will write equal like terms one side as:
$\Rightarrow 2+9=6x-x$
We can write the above equation as:
$\Rightarrow 6x-x=2+9$
Now, we will do the required calculations as:
$\Rightarrow 5x=11$
$\Rightarrow x=\dfrac{11}{5}$
Hence, the value of $x$ is $\dfrac{11}{5}$ .

Note: Here, we can check the solution is correct or not by applying the value of $x$ in the question that is:
$\Rightarrow 9\times {{3}^{x}}={{27}^{2x-3}}$
After putting the value of $x$ , we will have:
$\Rightarrow 9\times {{3}^{\dfrac{11}{5}}}={{27}^{2\times \dfrac{11}{5}-3}}$
Now, we will solve the equation of the power as:
$\Rightarrow {{3}^{2}}\times {{3}^{\dfrac{11}{5}}}={{27}^{\dfrac{22}{5}-3}}$
We can write the above step also as:
$\Rightarrow {{3}^{2+\dfrac{11}{5}}}={{27}^{\dfrac{22}{5}-3}}$
Here, we will use fraction addition method in the power one side and fraction subtraction method in the power other side as:
$\Rightarrow {{3}^{\dfrac{10+11}{5}}}={{27}^{\dfrac{22-15}{5}}}$
$\Rightarrow {{3}^{\dfrac{21}{5}}}={{27}^{\dfrac{7}{5}}}$
Since, $27$ can be written ${{3}^{3}}$ . Thus, the above equation will be as:
$\Rightarrow {{3}^{\dfrac{21}{5}}}={{\left( {{3}^{3}} \right)}^{\dfrac{7}{5}}}$
Now, we will open the bracket as:
$\Rightarrow {{3}^{\dfrac{21}{5}}}={{3}^{3\times \dfrac{7}{5}}}$
After completing the process we will get as:
$\Rightarrow {{3}^{\dfrac{21}{5}}}={{3}^{\dfrac{21}{5}}}$
Since, $L.H.S.=R.H.S.$
Hence, the solution is correct.