
If \[{{\log }_{8}}p=2.5\] and \[{{\log }_{2}}q=5\] , then p is expressed in terms of q is
\[\begin{align}
& \left( A \right)p={{q}^{2}} \\
& \left( B \right)p={{q}^{8}} \\
& \left( C \right)p={{q}^{\dfrac{3}{2}}} \\
& \left( D \right)p={{q}^{\dfrac{5}{2}}} \\
& \left( E \right)p={{q}^{3}} \\
\end{align}\]
Answer
542.7k+ views
Hint: These types of problems are pretty straight forward and are very simple to solve. Solving this problem requires some basic and advanced knowledge of logarithms as well as indices. We need to know about the basic operations and theorems of logarithms and transforming them into values using the power law. The general representation for the logarithm of a number is shown as,
\[{{\log }_{a}}b=c\] , Here a, is known as the base, b, is known as the exponent and c, is known as the value of the given logarithm. The basic form of operation of a logarithm is given by,
\[\begin{align}
& {{\log }_{a}}b=c \\
& \Rightarrow b={{a}^{c}} \\
\end{align}\]
We need to apply this same concept in our problem given here.
Complete step by step solution:
Now we start off with the solution to the given problem by writing that,
\[\begin{align}
& {{\log }_{8}}p=2.5 \\
& \Rightarrow p={{8}^{2.5}} \\
\end{align}\]
Now, we do this same thing to the other part that is given in the problem, i.e.
\[\begin{align}
& {{\log }_{2}}q=5 \\
& \Rightarrow q={{2}^{5}} \\
\end{align}\]
Now in the first equation we can see that, we can modify it further so that we can write the resultant equation in the form of the second equation. In this modification we just take the second equation and replace the value of \[8\] as \[{{2}^{3}}\]. Doing this will give us our required answer.
Thus,
\[\begin{align}
& p={{8}^{2.5}} \\
& \Rightarrow p={{2}^{3\times 2.5}} \\
& \Rightarrow p={{2}^{3\times \dfrac{5}{2}}} \\
& \Rightarrow p={{2}^{5\times \dfrac{3}{2}}} \\
\end{align}\]
We know from our first equation that \[q={{2}^{5}}\], Now replacing this in our second equation we get,
\[p={{q}^{\dfrac{3}{2}}}\]
So, the correct answer is “Option C”.
Note: For such problems we need to be very fluent with chapters and theory of logarithms and its various applications. We need to have a clear cut idea and understanding of the different theorems and laws of log. We must also be careful while conversion from the logarithmic form to the indices form so as to make the problem a bit easier. The rules of multiplication and division of logarithms come handy at times.
\[{{\log }_{a}}b=c\] , Here a, is known as the base, b, is known as the exponent and c, is known as the value of the given logarithm. The basic form of operation of a logarithm is given by,
\[\begin{align}
& {{\log }_{a}}b=c \\
& \Rightarrow b={{a}^{c}} \\
\end{align}\]
We need to apply this same concept in our problem given here.
Complete step by step solution:
Now we start off with the solution to the given problem by writing that,
\[\begin{align}
& {{\log }_{8}}p=2.5 \\
& \Rightarrow p={{8}^{2.5}} \\
\end{align}\]
Now, we do this same thing to the other part that is given in the problem, i.e.
\[\begin{align}
& {{\log }_{2}}q=5 \\
& \Rightarrow q={{2}^{5}} \\
\end{align}\]
Now in the first equation we can see that, we can modify it further so that we can write the resultant equation in the form of the second equation. In this modification we just take the second equation and replace the value of \[8\] as \[{{2}^{3}}\]. Doing this will give us our required answer.
Thus,
\[\begin{align}
& p={{8}^{2.5}} \\
& \Rightarrow p={{2}^{3\times 2.5}} \\
& \Rightarrow p={{2}^{3\times \dfrac{5}{2}}} \\
& \Rightarrow p={{2}^{5\times \dfrac{3}{2}}} \\
\end{align}\]
We know from our first equation that \[q={{2}^{5}}\], Now replacing this in our second equation we get,
\[p={{q}^{\dfrac{3}{2}}}\]
So, the correct answer is “Option C”.
Note: For such problems we need to be very fluent with chapters and theory of logarithms and its various applications. We need to have a clear cut idea and understanding of the different theorems and laws of log. We must also be careful while conversion from the logarithmic form to the indices form so as to make the problem a bit easier. The rules of multiplication and division of logarithms come handy at times.
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