Question

Evaluate ${8^{\dfrac{1}{3} + {{\log }_2}{{(121)}^{\dfrac{1}{3}}}}}$

Answer 1

In this sum the student has to use the properties of the logarithms which are $\log (ab) = \log a + \log b$,$\log (\dfrac{a}{b}) = \log a - \log b$,${\log _a}b = \dfrac{{\log b}}{{\log a}}$. The student has to use these properties one after the other. First step is using simplification. After that the student has to take the LCM and simplify the fraction before using the next property. In order to solve all the numericals related to logarithms the student needs to learn all the properties.

Complete step-by-step answer:
Let us assume that the sum is equal to a variable$x$. This variable has no meaning and it is redundant. We just have to keep it in order to apply the logarithmic properties while solving.
$\therefore x = {8^{\dfrac{1}{3} + {{\log }_2}{{(121)}^{\dfrac{1}{3}}}}}$
We can use the property of logarithms that is $\log {b^a} = a\log b$, in order to simplify the sum .
$\therefore x = {8^{\dfrac{1}{3} + \dfrac{1}{3}{{\log }_2}(121)}}$
We can take $\dfrac{1}{3}$common and then take the LCm of the remaining terms to solve the problem further.
$\therefore x = {8^{\dfrac{1}{3}(1 + {{\log }_2}(121))}}$
We can now use the property of logarithms which is ${\log _a}b = \dfrac{{\log b}}{{\log a}}$,thus after applying this we get the following step
$\therefore x = {8^{\dfrac{1}{3}(1 + \dfrac{{\log 121}}{{\log 2}})}}$
Taking LCM of the fraction we get the following step
$\therefore x = {8^{\dfrac{1}{3}(\dfrac{{\log 2 + \log 121}}{{\log 2}})}}$
Taking Log on both the sides we get the following equation
$\log x = \dfrac{1}{3}(\dfrac{{\log 2 + \log 121}}{{\log 2}})\log 8.......(1)$
We can also say that $\log 8$ is the same as $3\log 2$.
Thus substituting this value of $\log 8$ in equation $1$.
\Rightarrow $$$\log x = \dfrac{1}{3}(\dfrac{{\log 2 + \log 121}}{{\log 2}}) \times 3\log 2.......(2)$ Cancelling out the common terms we get the following equation \Rightarrow \log x = (\log 2 + \log 121).......(3)$ Using the property of Logs that i.e.$\log (a \times b) = \log a + \log b$ in the above equation we get next equation as follows $$ \Rightarrow \log x = \log (242).......(4)$Comparing Logs on both sides we can say that the value of$x$is$242$.

Thus , the value of the given sum is $242$.

Note: Only way to solve these numericals is to learn the properties and then apply them step by step as and when necessary. Sometimes students may get confused with the properties and might make mistakes while applying them. To prevent this from happening it is necessary that the student first notes down the property to be used in the sum in rough and then proceeds. Instead of blindly simplifying the sum sometimes it is necessary to find a common factor and strike it off before simplifying.