How do you solve \[{{\log }_{6}}x+{{\log }_{6}}3=2\]?
Answer
573.3k+ views
Hint:In the given question, we have been asked to find the value of ‘x’ and it is given that \[{{\log }_{6}}x+{{\log }_{6}}3=2\]. In order to find the value of ‘x’, first we will apply the law of logarithm which states that \[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\] . Then we need to apply the product property of logarithm which states that \[\log a+\log b=\log \left( a\times b \right)\] and simplify the equation further. After applying log formulae to the equation, we need to solve the equation in the way we solve general linear equations.
Complete step by step solution:
We have given,
\[{{\log }_{6}}x+{{\log }_{6}}3=2\]
Using the definition of logarithm, i.e.
\[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\]
Applying the definition of log, we get
\[\Rightarrow \dfrac{\log \left( x \right)}{\log \left( 6 \right)}+\dfrac{\log \left( 3 \right)}{\log \left( 6
\right)}=2\]
Multiply both the sides of the equation by log (7), we get
\[\Rightarrow \log 3+\log x=\log 32\]
Using the property of logarithm which states that if logs to the same base are added, then the numbers were multiplied, i.e. log (a) + log (b) = log (a.b)
\[\Rightarrow {{\log }_{6}}\left( 3x \right)=2\]
Using the definition of log,
If \[x\] and b are positive real numbers and b is not equal to 1,
Then \[{{\log }_{b}}\left( x \right)=y\]is equivalent to\[{{b}^{y}}=x\].
Therefore,
\[\Rightarrow 3x={{6}^{2}}\]
\[\Rightarrow 3x=36\]
Now solving for the value of ‘x’, we get
\[\Rightarrow x=\dfrac{36}{3}\]
\[\Rightarrow x=12\]
Therefore, the value of \[x=12\] is the required solution.
Formula used:
The definition of logarithm states that \[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\]
The property of logarithm which states that if logs to the same base are added, then the
numbers were multiplied, i.e. log (a) + log (b) = log (a.b).
Using the definition of log,
If \[x\] and b are positive real numbers and b is not equal to 1,
Then \[{{\log }_{b}}\left( x \right)=y\] is equivalent to \[{{b}^{y}}=x\].
Note: In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general linear equations.
Complete step by step solution:
We have given,
\[{{\log }_{6}}x+{{\log }_{6}}3=2\]
Using the definition of logarithm, i.e.
\[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\]
Applying the definition of log, we get
\[\Rightarrow \dfrac{\log \left( x \right)}{\log \left( 6 \right)}+\dfrac{\log \left( 3 \right)}{\log \left( 6
\right)}=2\]
Multiply both the sides of the equation by log (7), we get
\[\Rightarrow \log 3+\log x=\log 32\]
Using the property of logarithm which states that if logs to the same base are added, then the numbers were multiplied, i.e. log (a) + log (b) = log (a.b)
\[\Rightarrow {{\log }_{6}}\left( 3x \right)=2\]
Using the definition of log,
If \[x\] and b are positive real numbers and b is not equal to 1,
Then \[{{\log }_{b}}\left( x \right)=y\]is equivalent to\[{{b}^{y}}=x\].
Therefore,
\[\Rightarrow 3x={{6}^{2}}\]
\[\Rightarrow 3x=36\]
Now solving for the value of ‘x’, we get
\[\Rightarrow x=\dfrac{36}{3}\]
\[\Rightarrow x=12\]
Therefore, the value of \[x=12\] is the required solution.
Formula used:
The definition of logarithm states that \[{{\log }_{a}}x=\dfrac{{{\log }_{b}}x}{{{\log }_{b}}a}\]
The property of logarithm which states that if logs to the same base are added, then the
numbers were multiplied, i.e. log (a) + log (b) = log (a.b).
Using the definition of log,
If \[x\] and b are positive real numbers and b is not equal to 1,
Then \[{{\log }_{b}}\left( x \right)=y\] is equivalent to \[{{b}^{y}}=x\].
Note: In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general linear equations.
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