Answer
Verified
424.5k+ views
Hint: Solve the expression by using the laws of Logarithm. Take LHS and RHS of the expression separately. First the values by using the addition and exponential laws of logarithm. Simplify the final expression and find the value of x.
Complete step-by-step answer:
Let us consider that \[\log \left( x+5 \right)\] and \[\log \left( x-5 \right)\] are greater than zero.
i.e. \[\log \left( x+5 \right)>0\] and \[\log \left( x-5 \right)>0\].
We have been given the expression, \[\log \left( x+5 \right)+\log \left( x-5 \right)=4\log 2+2\log 3\].
By using the laws of logarithm we can solve this expression.
\[\log A+\log B=\log AB-(1)\]
This expression tells us how to add two logarithms. Adding \[\log A\] and \[\log B\] results in the logarithm of the product A and B, that is \[\log AB\].
\[\log {{A}^{n}}=n\log A-(2)\]
In this case the logarithm can be written in exponential form.
Now consider the RHS of the expression,
\[\log \left( x+5 \right)+\log \left( x-5 \right)\]
From equation (1) we can say that \[\log A+\log B=\log AB\].
Comparing both the expression, we get value of
A = x+5 and B = x- 5
Now substituting value of A and B in equation (1),
\[\log \left( x+5 \right)+\log \left( x-5 \right)=\log \left( \left( x+5 \right)\left( x-5 \right) \right)-(3)\]
Now let us consider the LHS of the expression \[\left( 4\log 2+2\log 3 \right)\].
From Equation (2), we know, \[x\log A=\log {{A}^{x}}\].
\[\therefore 4\log 2\] can be written as log\[{{2}^{4}}\].
Similarly, \[2\log 3\] can be written as \[\log {{3}^{2}}\].
Now, \[\left( 4\log 2+2\log 3 \right)\] becomes \[\left( \log {{2}^{4}}+\log {{3}^{2}} \right)\].
From equation (1) we know that \[\log A+\log B=\log AB\]. Comparing both expressions we get \[A={{2}^{4}}\] and \[B={{3}^{2}}\]
Substituting, the values of A and B we get,
\[\log {{2}^{4}}+\log {{3}^{2}}=\log \left( {{2}^{4}}\times {{3}^{2}} \right)-(4)\]
Now let us equate the RHS and LHS of the expression.
We get,
\[\log \left( x+5 \right)\left( x-5 \right)=\log \left( {{2}^{4}}\times {{3}^{2}} \right)\]
Let us remove the log from both sides.
\[\Rightarrow \left( x+5 \right)\left( x-5 \right)={{2}^{4}}\times {{3}^{2}}\]
Simplifying and opening the brackets,
\[\begin{align}
& \left( x+5 \right)\left( x-5 \right)={{x}^{2}}-5x+5x-25={{x}^{2}}-25 \\
& {{2}^{4}}=2\times 2\times 2\times 2=16 \\
& {{3}^{2}}=3\times 3=9 \\
& \Rightarrow {{x}^{2}}-25=16\times 9 \\
& {{x}^{2}}=144+25=169 \\
& \therefore {{x}^{2}}=169 \\
& x=\sqrt{169}=13 \\
\end{align}\]
\[\therefore \] Value of \[x=\pm 13\]
Since, we can’t consider x = -13 due to the fact that the argument of any logarithm can’t be negative.
\[\therefore \] Option (d) is the correct answer.
Note: The given laws of logarithms are first law and third laws of the logarithm.
The second law is \[\log A-\log B=\log \dfrac{A}{B}\] (subtraction of logarithms).
And the fourth law is \[\log 1=0,{{\log }_{m}}m=1\].
The logarithm of 1 to any base is always zero, and the logarithm of a number to the same bare is always 1.
\[{{\log }_{10}}10=1\]and \[{{\log }_{e}}e=1\].
The law of logarithm applies to the logarithm of any base. But the same bare should be used throughout the calculation.
Complete step-by-step answer:
Let us consider that \[\log \left( x+5 \right)\] and \[\log \left( x-5 \right)\] are greater than zero.
i.e. \[\log \left( x+5 \right)>0\] and \[\log \left( x-5 \right)>0\].
We have been given the expression, \[\log \left( x+5 \right)+\log \left( x-5 \right)=4\log 2+2\log 3\].
By using the laws of logarithm we can solve this expression.
\[\log A+\log B=\log AB-(1)\]
This expression tells us how to add two logarithms. Adding \[\log A\] and \[\log B\] results in the logarithm of the product A and B, that is \[\log AB\].
\[\log {{A}^{n}}=n\log A-(2)\]
In this case the logarithm can be written in exponential form.
Now consider the RHS of the expression,
\[\log \left( x+5 \right)+\log \left( x-5 \right)\]
From equation (1) we can say that \[\log A+\log B=\log AB\].
Comparing both the expression, we get value of
A = x+5 and B = x- 5
Now substituting value of A and B in equation (1),
\[\log \left( x+5 \right)+\log \left( x-5 \right)=\log \left( \left( x+5 \right)\left( x-5 \right) \right)-(3)\]
Now let us consider the LHS of the expression \[\left( 4\log 2+2\log 3 \right)\].
From Equation (2), we know, \[x\log A=\log {{A}^{x}}\].
\[\therefore 4\log 2\] can be written as log\[{{2}^{4}}\].
Similarly, \[2\log 3\] can be written as \[\log {{3}^{2}}\].
Now, \[\left( 4\log 2+2\log 3 \right)\] becomes \[\left( \log {{2}^{4}}+\log {{3}^{2}} \right)\].
From equation (1) we know that \[\log A+\log B=\log AB\]. Comparing both expressions we get \[A={{2}^{4}}\] and \[B={{3}^{2}}\]
Substituting, the values of A and B we get,
\[\log {{2}^{4}}+\log {{3}^{2}}=\log \left( {{2}^{4}}\times {{3}^{2}} \right)-(4)\]
Now let us equate the RHS and LHS of the expression.
We get,
\[\log \left( x+5 \right)\left( x-5 \right)=\log \left( {{2}^{4}}\times {{3}^{2}} \right)\]
Let us remove the log from both sides.
\[\Rightarrow \left( x+5 \right)\left( x-5 \right)={{2}^{4}}\times {{3}^{2}}\]
Simplifying and opening the brackets,
\[\begin{align}
& \left( x+5 \right)\left( x-5 \right)={{x}^{2}}-5x+5x-25={{x}^{2}}-25 \\
& {{2}^{4}}=2\times 2\times 2\times 2=16 \\
& {{3}^{2}}=3\times 3=9 \\
& \Rightarrow {{x}^{2}}-25=16\times 9 \\
& {{x}^{2}}=144+25=169 \\
& \therefore {{x}^{2}}=169 \\
& x=\sqrt{169}=13 \\
\end{align}\]
\[\therefore \] Value of \[x=\pm 13\]
Since, we can’t consider x = -13 due to the fact that the argument of any logarithm can’t be negative.
\[\therefore \] Option (d) is the correct answer.
Note: The given laws of logarithms are first law and third laws of the logarithm.
The second law is \[\log A-\log B=\log \dfrac{A}{B}\] (subtraction of logarithms).
And the fourth law is \[\log 1=0,{{\log }_{m}}m=1\].
The logarithm of 1 to any base is always zero, and the logarithm of a number to the same bare is always 1.
\[{{\log }_{10}}10=1\]and \[{{\log }_{e}}e=1\].
The law of logarithm applies to the logarithm of any base. But the same bare should be used throughout the calculation.
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE