
Following is simplifies to: ${{\log }_{10}}\left( {{\log }_{2}}3 \right)+{{\log }_{10}}\left( {{\log }_{3}}4 \right)+{{\log }_{10}}\left( {{\log }_{4}}5 \right)+\ldots \ldots +{{\log }_{10}}\left( {{\log }_{1023}}1024 \right)$
A. A composite number.
B. A prime number.
C. Rational number which is not an integer.
D. An integer number.
Answer
489.6k+ views
Hint: For solving this question you should know about the addition of logarithmic terms if their base is changing continuously. IN this problem first we will write down the expression and then start solving this by taking ${{\log }_{10}}$ common and rest all the terms will be in multiplication form. So, the terms which are the same and, in the division, cancel to each other and at last which term will remain, we will solve for that and find the answer.
Complete step by step answer:
So, according to the question an expression is given as in logarithmic term form and we have to simplify that. As we know that,
${{\log }_{a}}b=\dfrac{\log b}{\log a}$
And,
${{\log }_{a}}+{{\log }_{b}}=\log \left( a.b \right)$
Now given : ${{\log }_{10}}\left( {{\log }_{2}}3 \right)+{{\log }_{10}}\left( {{\log }_{3}}4 \right)+{{\log }_{10}}\left( {{\log }_{4}}5 \right)+\ldots \ldots +{{\log }_{10}}\left( {{\log }_{1023}}1024 \right)$.
So, if we solve this with the help of these formulas, then we can write it as,
$\begin{align}
& {{\log }_{10}}\left( {{\log }_{2}}3 \right)+{{\log }_{10}}\left( {{\log }_{3}}4 \right)+{{\log }_{10}}\left( {{\log }_{4}}5 \right)+\ldots +{{\log }_{10}}\left( {{\log }_{1023}}1024 \right) \\
& ={{\log }_{10}}\left( \dfrac{\log 3}{\log 2}.\dfrac{\log 4}{\log 3}.\dfrac{\log 5}{\log 4}.\dfrac{\log 6}{\log 5}\ldots \dfrac{\log 1024}{\log 1023} \right) \\
\end{align}$
And if we solve this, then we get,
$\begin{align}
& \Rightarrow {{\log }_{10}}\left( \dfrac{\log 1024}{\log 2} \right)=\log \left( \dfrac{\log {{2}^{10}}}{\log 2} \right) \\
& \Rightarrow {{\log }_{10}}\left( \dfrac{\log {{2}^{10}}}{\log 2} \right)={{\log }_{10}}\left( 10 \right)=1 \\
\end{align}$
So, the correct answer is “Option D”.
Note: While solving these types of questions you have to keep in mind that the log has a general property for making log’s long terms in short terms. We have to always keep in mind all the formulas of the log and then solve the question. And always try to reduce the equation as much as possible because it will provide you with an accurate answer.
Complete step by step answer:
So, according to the question an expression is given as in logarithmic term form and we have to simplify that. As we know that,
${{\log }_{a}}b=\dfrac{\log b}{\log a}$
And,
${{\log }_{a}}+{{\log }_{b}}=\log \left( a.b \right)$
Now given : ${{\log }_{10}}\left( {{\log }_{2}}3 \right)+{{\log }_{10}}\left( {{\log }_{3}}4 \right)+{{\log }_{10}}\left( {{\log }_{4}}5 \right)+\ldots \ldots +{{\log }_{10}}\left( {{\log }_{1023}}1024 \right)$.
So, if we solve this with the help of these formulas, then we can write it as,
$\begin{align}
& {{\log }_{10}}\left( {{\log }_{2}}3 \right)+{{\log }_{10}}\left( {{\log }_{3}}4 \right)+{{\log }_{10}}\left( {{\log }_{4}}5 \right)+\ldots +{{\log }_{10}}\left( {{\log }_{1023}}1024 \right) \\
& ={{\log }_{10}}\left( \dfrac{\log 3}{\log 2}.\dfrac{\log 4}{\log 3}.\dfrac{\log 5}{\log 4}.\dfrac{\log 6}{\log 5}\ldots \dfrac{\log 1024}{\log 1023} \right) \\
\end{align}$
And if we solve this, then we get,
$\begin{align}
& \Rightarrow {{\log }_{10}}\left( \dfrac{\log 1024}{\log 2} \right)=\log \left( \dfrac{\log {{2}^{10}}}{\log 2} \right) \\
& \Rightarrow {{\log }_{10}}\left( \dfrac{\log {{2}^{10}}}{\log 2} \right)={{\log }_{10}}\left( 10 \right)=1 \\
\end{align}$
So, the correct answer is “Option D”.
Note: While solving these types of questions you have to keep in mind that the log has a general property for making log’s long terms in short terms. We have to always keep in mind all the formulas of the log and then solve the question. And always try to reduce the equation as much as possible because it will provide you with an accurate answer.
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