Show that xyz=2. If \[x={{\log }_{7}}9\], $y={{\log }_{5}}7$, $z={{\log }_{3}}5$.
Answer
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Hint: For solving this question first we will write $x$, $y$ and $z$ into fraction form with the help of some logarithmic identities. Then we will multiply them easily to prove the result.
Complete step-by-step answer:
Given:
It is given that if \[x={{\log }_{7}}9\] , $y={{\log }_{5}}7$, $z={{\log }_{3}}5$ and we have to prove that $xyz=2$.
Now, before we proceed we should be familiar with the following logarithmic formulas:
1. If $a$ and $b$ are two positive numbers then, ${{\log }_{b}}a=\dfrac{{{\log }_{10}}a}{{{\log }_{10}}b}=\dfrac{\log a}{\log b}$.
2. If $a$ is any positive number then, $\log {{a}^{n}}=n\log a$.
Now, we will use the above-mentioned formulas to write $x$, $y$ and $z$ into some other form that will help us to find the value of $xyz$ easily.
Now, as it is given that \[x={{\log }_{7}}9\] , using the formula mentioned in the first and second point. Then,
\[\begin{align}
& x={{\log }_{7}}9 \\
& \Rightarrow x=\dfrac{\log 9}{\log 7}=\dfrac{\log {{3}^{2}}}{\log 7} \\
& \Rightarrow x=\dfrac{2\log 3}{\log 7}..................\left( 1 \right) \\
\end{align}\]
Now, as it is given that $y={{\log }_{5}}7$ , using the formula mentioned in the first point. Then,
$\begin{align}
& y={{\log }_{5}}7 \\
& \Rightarrow y=\dfrac{\log 7}{\log 5}..............\left( 2 \right) \\
\end{align}$
Now, as it is given that $z={{\log }_{3}}5$ , using the formula mentioned in the first point. Then,
$\begin{align}
& z={{\log }_{3}}5 \\
& \Rightarrow z=\dfrac{\log 5}{\log 3}................\left( 3 \right) \\
\end{align}$
Now, from equation (1) we have \[x=\dfrac{2\log 3}{\log 7}\] , from equation (2) we have $y=\dfrac{\log 7}{\log 5}$ and from equation (3) we have $z=\dfrac{\log 5}{\log 3}$ . Then,
$\begin{align}
& xyz=\dfrac{2\log 3}{\log 7}\times \dfrac{\log 7}{\log 5}\times \dfrac{\log 5}{\log 3} \\
& \Rightarrow xyz=2 \\
\end{align}$
Thus, the value of $xyz=2$.
Hence, proved.
Note: Here, the student should avoid multiplying the given terms directly without using logarithmic identities. Firstly, the value of given terms should be written in such a manner so that when we multiply them there won’t be any difficulty. Moreover, students should avoid calculation mistakes while solving the problem to get the correct answer quickly.
Complete step-by-step answer:
Given:
It is given that if \[x={{\log }_{7}}9\] , $y={{\log }_{5}}7$, $z={{\log }_{3}}5$ and we have to prove that $xyz=2$.
Now, before we proceed we should be familiar with the following logarithmic formulas:
1. If $a$ and $b$ are two positive numbers then, ${{\log }_{b}}a=\dfrac{{{\log }_{10}}a}{{{\log }_{10}}b}=\dfrac{\log a}{\log b}$.
2. If $a$ is any positive number then, $\log {{a}^{n}}=n\log a$.
Now, we will use the above-mentioned formulas to write $x$, $y$ and $z$ into some other form that will help us to find the value of $xyz$ easily.
Now, as it is given that \[x={{\log }_{7}}9\] , using the formula mentioned in the first and second point. Then,
\[\begin{align}
& x={{\log }_{7}}9 \\
& \Rightarrow x=\dfrac{\log 9}{\log 7}=\dfrac{\log {{3}^{2}}}{\log 7} \\
& \Rightarrow x=\dfrac{2\log 3}{\log 7}..................\left( 1 \right) \\
\end{align}\]
Now, as it is given that $y={{\log }_{5}}7$ , using the formula mentioned in the first point. Then,
$\begin{align}
& y={{\log }_{5}}7 \\
& \Rightarrow y=\dfrac{\log 7}{\log 5}..............\left( 2 \right) \\
\end{align}$
Now, as it is given that $z={{\log }_{3}}5$ , using the formula mentioned in the first point. Then,
$\begin{align}
& z={{\log }_{3}}5 \\
& \Rightarrow z=\dfrac{\log 5}{\log 3}................\left( 3 \right) \\
\end{align}$
Now, from equation (1) we have \[x=\dfrac{2\log 3}{\log 7}\] , from equation (2) we have $y=\dfrac{\log 7}{\log 5}$ and from equation (3) we have $z=\dfrac{\log 5}{\log 3}$ . Then,
$\begin{align}
& xyz=\dfrac{2\log 3}{\log 7}\times \dfrac{\log 7}{\log 5}\times \dfrac{\log 5}{\log 3} \\
& \Rightarrow xyz=2 \\
\end{align}$
Thus, the value of $xyz=2$.
Hence, proved.
Note: Here, the student should avoid multiplying the given terms directly without using logarithmic identities. Firstly, the value of given terms should be written in such a manner so that when we multiply them there won’t be any difficulty. Moreover, students should avoid calculation mistakes while solving the problem to get the correct answer quickly.
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