How do you find\[\log {{x}^{2}}+\log 25=2\]?
Answer
619.5k+ views
Hint:In the given question, we have been asked to find the value of ‘x’ in \[\log {{x}^{2}}+\log 25=2\]. In order to find the value of ‘x’, first we need to use the product property of logarithm i.e. \[{{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right)={{\log }_{b}}\left( xy \right)\]. Later we solve the equation by using the definition of logarithm i.e. 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\[{{b}^{y}}=x\]. And then simplifying the question using mathematical operations such as addition, subtraction, multiplication and division.
Formula used:
The logarithm product property,
\[{{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right)={{\log }_{b}}\left( xy \right)\]
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\[{{b}^{y}}=x\].
Complete step by step solution:
We have given,
\[\Rightarrow \]\[\log {{x}^{2}}+\log 25=2\]
Using the logarithm product property,
\[{{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right)={{\log }_{b}}\left( xy \right)\]
Simplifying the left side by using product property of algorithm, we get
\[\Rightarrow \log \left( {{x}_{2}}\times 25 \right)=2\]
\[\Rightarrow \log \left( 25{{x}^{2}} \right)=2\]
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,
\[\Rightarrow {{10}^{2}}=25{{x}^{2}}\]
Rearranging the equation, we get
\[\Rightarrow 25{{x}^{2}}={{10}^{2}}\]
Simplifying the above, we get
\[\Rightarrow 25{{x}^{2}}=100\]
Dividing both the sides of the equation, we get
\[\Rightarrow {{x}^{2}}=4\]
Solving for the value of ‘x’, we get
\[\Rightarrow x=\sqrt{4}=\pm 2\]
Therefore,
\[\Rightarrow x=2,-2\]
Hence, the values of ‘x’ are 2 and -2 is the required solution.
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 be required 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.
Formula used:
The logarithm product property,
\[{{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right)={{\log }_{b}}\left( xy \right)\]
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\[{{b}^{y}}=x\].
Complete step by step solution:
We have given,
\[\Rightarrow \]\[\log {{x}^{2}}+\log 25=2\]
Using the logarithm product property,
\[{{\log }_{b}}\left( x \right)+{{\log }_{b}}\left( y \right)={{\log }_{b}}\left( xy \right)\]
Simplifying the left side by using product property of algorithm, we get
\[\Rightarrow \log \left( {{x}_{2}}\times 25 \right)=2\]
\[\Rightarrow \log \left( 25{{x}^{2}} \right)=2\]
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,
\[\Rightarrow {{10}^{2}}=25{{x}^{2}}\]
Rearranging the equation, we get
\[\Rightarrow 25{{x}^{2}}={{10}^{2}}\]
Simplifying the above, we get
\[\Rightarrow 25{{x}^{2}}=100\]
Dividing both the sides of the equation, we get
\[\Rightarrow {{x}^{2}}=4\]
Solving for the value of ‘x’, we get
\[\Rightarrow x=\sqrt{4}=\pm 2\]
Therefore,
\[\Rightarrow x=2,-2\]
Hence, the values of ‘x’ are 2 and -2 is the required solution.
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 be required 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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