
If three terms ${{\log }_{x}}2,{{2}^{\dfrac{x}{2}}},{{\log }_{3}}x$ are in G.P., then the value of x equals to –
A). \[{{\log }_{3}}({{\log }_{3}}2)\]
B). \[{{\log }_{3}}({{\log }_{2}}3)\]
C). \[{{\log }_{2}}({{\log }_{3}}2)\]
D). \[{{\log }_{2}}({{\log }_{2}}3)\]
Answer
581.1k+ views
Hint: In the above question we will use the concept of G.P as well as the properties of a logarithmic function. A geometric progression (G.P), also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non- zero number called the common ratio. The relation of (G.P) that we will use is as follow,
\[{{b}^{2}}=a\times c\] where a, b and c are the consecutive terms of a G.P.
Also, we will use the formula of a logarithmic as shown below,
\[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\]
Complete-step-by-step solution
Now, we have been given that ${{\log }_{x}}2,{{2}^{\dfrac{x}{2}}},{{\log }_{3}}x$ are in G.P, so we will use the formula of G.P \[{{b}^{2}}=a\times c\] and we get,
\[\Rightarrow {{\left( {{2}^{\dfrac{x}{2}}} \right)}^{2}}={{\log }_{x}}2\times {{\log }_{3}}x\]
Now, we can further simplify using the formula \[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\] as,
\[\begin{align}
& \Rightarrow {{2}^{\dfrac{x}{2}\times 2}}=\dfrac{\log 2}{\log x}\times \dfrac{\log x}{\log 3} \\
& \Rightarrow {{2}^{x}}={{\log }_{3}}2 \\
\end{align}\]
Now, taking ${{\log }_{2}}$ on both sides of the above equation, we get
\[\Rightarrow {{\log }_{2}}\left( {{2}^{x}} \right)={{\log }_{2}}({{\log }_{3}}2)\]
Since we know that ${{\log }_{e}}{{a}^{b}}=b{{\log }_{e}}a$, we can write the term on LHS as,
\[\begin{align}
& \Rightarrow x\times {{\log }_{2}}2={{\log }_{2}}({{\log }_{3}}2) \\
& \Rightarrow x={{\log }_{2}}({{\log }_{3}}2) \\
\end{align}\]
Therefore the correct option of the above question is option C.
Note: Just remember the formula for the G.P and logarithmic and also be careful while doing calculation as there is a chance that you might make silly mistakes and you will get the incorrect answer. Just go once through all the properties of logarithmic function as well as the properties related to G.P as it will help you a lot in these types of problems.
\[{{b}^{2}}=a\times c\] where a, b and c are the consecutive terms of a G.P.
Also, we will use the formula of a logarithmic as shown below,
\[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\]
Complete-step-by-step solution
Now, we have been given that ${{\log }_{x}}2,{{2}^{\dfrac{x}{2}}},{{\log }_{3}}x$ are in G.P, so we will use the formula of G.P \[{{b}^{2}}=a\times c\] and we get,
\[\Rightarrow {{\left( {{2}^{\dfrac{x}{2}}} \right)}^{2}}={{\log }_{x}}2\times {{\log }_{3}}x\]
Now, we can further simplify using the formula \[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\] as,
\[\begin{align}
& \Rightarrow {{2}^{\dfrac{x}{2}\times 2}}=\dfrac{\log 2}{\log x}\times \dfrac{\log x}{\log 3} \\
& \Rightarrow {{2}^{x}}={{\log }_{3}}2 \\
\end{align}\]
Now, taking ${{\log }_{2}}$ on both sides of the above equation, we get
\[\Rightarrow {{\log }_{2}}\left( {{2}^{x}} \right)={{\log }_{2}}({{\log }_{3}}2)\]
Since we know that ${{\log }_{e}}{{a}^{b}}=b{{\log }_{e}}a$, we can write the term on LHS as,
\[\begin{align}
& \Rightarrow x\times {{\log }_{2}}2={{\log }_{2}}({{\log }_{3}}2) \\
& \Rightarrow x={{\log }_{2}}({{\log }_{3}}2) \\
\end{align}\]
Therefore the correct option of the above question is option C.
Note: Just remember the formula for the G.P and logarithmic and also be careful while doing calculation as there is a chance that you might make silly mistakes and you will get the incorrect answer. Just go once through all the properties of logarithmic function as well as the properties related to G.P as it will help you a lot in these types of problems.
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