Question

If ${x^5} - 8 = 159$, what is the approximate value of $x$?
A. $2.67$
B. $2.71$
C. $2.78$
D. $2.81$
E. $2.84$

Answer 1

In this problem, first we will rewrite the given equation. Then, we will use differentials to approximate the required value of $x$. We know that the increment in variable $y$ corresponding to the increment in variable $x$ is given by $\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)$. Also we know that the differential of $x$ is defined as $dx = \Delta x$ and the differential of $y$ is defined as $dy = f'\left( x \right)dx$ or $dy = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x$. We will use this information to approximate the value of $x$.

Complete step-by-step answer:
In this problem, the given equation is ${x^5} - 8 = 159$. Let us rewrite this equation. Therefore, we get ${x^5} = 159 + 8 \\\ \Rightarrow {x^5} = 167 \\\ \Rightarrow x = {\left( {167} \right)^{\dfrac{1}{5}}}\; \cdots \cdots \left( 1 \right) \\\$
Now we need to find the fifth root of the number $167$. First we will think about the number whose fifth root is a positive integer. We know that ${\left( {243} \right)^{\dfrac{1}{5}}} = 3$. Therefore, we will rewrite the equation $\left( 1 \right)$ as ${\left( {167} \right)^{\dfrac{1}{5}}} = {\left( {243 - 76} \right)^{\dfrac{1}{5}}}$.
To approximate this value, let us consider $y = f\left( x \right) = {x^{\dfrac{1}{5}}}$ with $x = 243$ and $\Delta x = - 76$.
We know that the increment in variable $y$ corresponding to the increment in variable $x$ is given by $\Delta y = f\left( {x + \Delta x} \right) - f\left( x \right)$. Therefore, we can write
$\Delta y = {\left( {x + \Delta x} \right)^{\dfrac{1}{5}}} - {x^{\dfrac{1}{5}}}\quad \left[ {\because f\left( x \right) = {x^{\dfrac{1}{5}}}} \right] \\\ \Rightarrow \Delta y = {\left( {243 - 76} \right)^{\dfrac{1}{5}}} - {\left( {243} \right)^{\dfrac{1}{5}}}\quad \left[ {\because x = 243,\;\Delta x = - 76} \right] \\\ \Rightarrow \Delta y = {\left( {167} \right)^{\dfrac{1}{5}}} - 3 \\\ \Rightarrow {\left( {167} \right)^{\dfrac{1}{5}}} = 3 + \Delta y \cdots \cdots \left( 2 \right) \\\$
Also we know that the differential of $x$ is defined as $dx = \Delta x$ and the differential of $y$ is defined as $dy = f'\left( x \right)dx$ or $dy = \left( {\dfrac{{dy}}{{dx}}} \right)\Delta x$. Therefore, we can write
$dy = \left( {\dfrac{1}{5}{x^{\dfrac{1}{5} - 1}}} \right)\left( { - 76} \right) \cdots \cdots \left( 3 \right)\quad \left[ {\because y = {x^{\dfrac{1}{5}}},\;\Delta x = - 76} \right]$
Note the here we find $\dfrac{{dy}}{{dx}}$ using the formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$. Let us simplify the equation $\left( 3 \right)$. Therefore, we get $dy = - \dfrac{{76{x^{\left( { - \dfrac{4}{5}} \right)}}}}{5}$
$\Rightarrow dy = - \dfrac{{76{{\left( {243} \right)}^{ - \dfrac{4}{5}}}}}{5}\quad \left[ {\because x = 243} \right] \\\ \Rightarrow dy = - \dfrac{{76{{\left( {{3^5}} \right)}^{ - \dfrac{4}{5}}}}}{5} \\\ \Rightarrow dy = - \dfrac{{76{{\left( 3 \right)}^{ - 5 \times \dfrac{4}{5}}}}}{5}\quad \left[ {\because {{\left( {{a^m}} \right)}^n} = {a^{m\; \times \;n}}} \right] \\\ \Rightarrow dy = - \dfrac{{76{{\left( 3 \right)}^{ - 4}}}}{5} \\\ \Rightarrow dy = - \dfrac{{76}}{{5{{\left( 3 \right)}^4}}} \\\ \Rightarrow dy = - \dfrac{{76}}{{5\left( {81} \right)}} \\\ \Rightarrow dy = - 0.1877 \\\$
Note that here $dy$ is approximately equal to $\Delta y$. Therefore, $\Delta y = - 0.1877$.
Now we will put the value of $\Delta y$ in equation $\left( 2 \right)$, we get ${\left( {167} \right)^{\dfrac{1}{5}}} = 3 + \left( { - 0.1877} \right)$
$\Rightarrow {\left( {167} \right)^{\dfrac{1}{5}}} = 2.8123$. Therefore, if ${x^5} - 8 = 159$ then the approximate value of $x$ is $2.81$.
Therefore, option D is correct.

Note: In this type of problem if $dx = \Delta x$ is relatively small (when compared with $x$) then $dy$ will be a good approximation of $\Delta y$ and it is denoted by $dy \approx \Delta y$. Also note that the differential of the dependent variable (usually $y$) is not equal to the increment of the variable but the differential of the independent variable (usually $x$) is equal to the increment of variable. In this problem, $y$ is dependent variable and $x$ is independent variable.