Question

Solve $x={{125}^{\dfrac{-1}{3}}}$ and find the value of $10x$.

Answer 1

Hint: To solve this question, we will cube both sides and will find the cube root of the resulting number.

Complete step-by-step answer:
According to question, we are given that;
$x={{125}^{\dfrac{-1}{3}}}$
We will rearrange the term on the right-hand side. After rearranging, we will get following equation: -
$x={{125}^{\dfrac{-1}{3}}}$………………….(i)
We can write the above equation as follows: -
$x={{\left( 125 \right)}^{\dfrac{-1}{3}}}$ ……………………(ii)
We can write this in the following manner because we know that;
${{p}^{-1}}=\dfrac{1}{p}$
Also, we know that the cube root of 1 is 1itself, we can write ${{1}^{\dfrac{1}{3}}}=1$. Also, there is another identity which is being used when we have covert the equation (i) into equation (ii). Thus, identity is given as follows: -
$\dfrac{a''}{b''}={{\left( \dfrac{a}{b} \right)}^{''}}$

Now, we will continue from the equation (ii). We will now cube both the sides of the equation (ii). After cubing, we will get the following result: -
${{\left( x \right)}^{3}}={{\left[ {{125}^{\dfrac{-1}{3}}} \right]}^{3}}$ ………………………………..(iii)
Here, we will use exponential identity which is given as follows: -
${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$
Applying this identity to the equation (iii), we will get following result: -
$\Rightarrow {{x}^{3}}={{\left( \dfrac{1}{125} \right)}^{\dfrac{1}{3}\times }}^{3}$
$\Rightarrow {{x}^{3}}={{\left( \dfrac{1}{125} \right)}^{1}}$
$\Rightarrow {{x}^{3}}=\left( \dfrac{1}{125} \right)$
As we can see that in the left hand side, there is a cubic form, so we will have to create a cubic from in the right hand side also to do this we will write 1 as ${{1}^{3}}$ and 125 as ${{5}^{3}}$. After putting these values, we will get the following results: -
$\Rightarrow {{x}^{3}}=\dfrac{{{1}^{3}}}{{{5}^{3}}}$
This can also be written as follows
$\Rightarrow {{x}^{3}}={{\left( \dfrac{1}{5} \right)}^{3}}$
Now, we will compare the base and power with power. As the power on LHS and RHS is same, the base should also be same, i.e.
$\Rightarrow x=\dfrac{1}{5}$
$\Rightarrow x=0.2$
Now, as given in the equation, we have to find the value of $10x$.
$10x=10\times x=10\times 0.2$
$=10\times \dfrac{1}{5}=2$
So, the value of $10x=2$

Note: We could have solved the above equation by taking log on both sides with its base 5. This is as shown below: -
${{\log }_{5}}x={{\log }_{5}}{{125}^{\dfrac{-1}{3}}}=\dfrac{-1}{3}{{\log }_{5}}125$
$\Rightarrow {{\log }_{5}}x=\dfrac{-1}{3}{{\log }_{5}}{{5}^{3}}$
$\Rightarrow {{\log }_{5}}x=\dfrac{-1}{3}\times 3\times {{\log }_{5}}{{5}^{3}}=-1$
$\Rightarrow x={{5}^{-1}}=\dfrac{1}{5}$