Question

The approximate value of ${5^{2.01}}$ is _______. Where, $\left( {{{\log }_e}5 = 1.6095} \right)$
$
  A.{\text{ 25}}{\text{.4125}} \\
  B.{\text{ 25}}{\text{.2525}} \\
  C.{\text{ 25}}{\text{.5025}} \\
  D.{\text{ }}25.4024 \\
 $

Answer 1

Hint- In order to solve such a question first define a general function according to the problem. Use the concept of approximation. Then with the help of a derivative of the function, try to solve the problem.

Complete step-by-step solution-
Given value to be found out is ${5^{2.01}}$ also we have $\left( {{{\log }_e}5 = 1.6095} \right)$
We know the concept of approximation of derivatives. To find a very small change or variation of a quantity, we can use derivatives to give the approximate value of it. The approximate value is represented by delta.
\[\dfrac{{dy}}{{dx}} = \dfrac{{\left[ {f\left( {x + \Delta x} \right) - f\left( x \right)} \right]}}{{\left( {x + \Delta x} \right) - x}}\]
Let according to the problem the function is
$f\left( x \right) = {5^x},x = 2\& \Delta x = 0.01$
As we know by the general definition that
\[f\left( {x + \Delta x} \right) = f\left( x \right) + \dfrac{{dy}}{{dx}}.\Delta x\]
Substituting the values in above formula we get:
\[f\left( {2 + 0.01} \right) = f\left( 2 \right) + \dfrac{{d\left( {{5^x}} \right)}}{{dx}}.\left( {0.01} \right)\]
Now let us differentiate and proceed further
\[
   \Rightarrow f\left( {2 + 0.01} \right) = f\left( 2 \right) + {5^x}\ln \left( 5 \right).\left( {0.01} \right){\text{ }}\left[ {\because \dfrac{{d\left( {{a^x}} \right)}}{{dx}} = {a^x}\ln \left( a \right)} \right] \\
   \Rightarrow f\left( {2 + 0.01} \right) = {5^2} + {5^2}\ln \left( 5 \right).\left( {0.01} \right){\text{ }}\left[ {\because x = 2\& f\left( x \right) = {5^x}} \right] \\
   \Rightarrow {5^{2 + 0.01}} = 25 + \left( {25} \right)\left( {1.6095} \right)\left( {0.01} \right){\text{ }}\left[ {\because \ln \left( 5 \right) = 1.6095} \right] \\
   \Rightarrow {5^{2.01}} = 25 + 0.402375 \\
   \Rightarrow {5^{2.01}} = 25.402375 \\
 \]
Hence, the approximate value of ${5^{2.01}}$ is 25.4024.
So, option D is the correct option.

Note- These types of problems containing decimal power of any number cannot be found out by simple calculation so the differentiating method is one of the easiest ways to solve it. It can also be found out with the help of a logarithm table and slide rule. Iterative method of solution is another good way but that involves consumption of lots of time.