Answer

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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.

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