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Solve $x$ if $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{2-\sqrt{3}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

$x$ is a literal number and it formed an equation in terms of logarithms and it is required to find the value of the $x$ by solving this logarithmic equation mathematically.

$2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{2-\sqrt{3}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$


Making bases of both terms same

The bases of both logarithmic terms are different. Hence, it is not possible to simplify it but it can be simplified by making both bases same. Rationalising method is mostly useful to do it.

$\implies$ $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{(2-\sqrt{3}) \times \Large \frac{2+\sqrt{3}}{2+\sqrt{3}}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

$\implies$ $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{\Large \frac{(2-\sqrt{3})(2+\sqrt{3})}{2+\sqrt{3}}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

$\implies$ $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{\Large \frac{2^2-{(\sqrt{3})}^2}{2+\sqrt{3}}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

$\implies$ $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{\Large \frac{4-3}{2+\sqrt{3}}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

$\implies$ $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{\Large \frac{1}{2+\sqrt{3}}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

Try negative exponent law to move one step in making the base of the second term same as the first term.

$\implies$ $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{{(2+\sqrt{3})}^{\Large -1}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

Apply, power rule of logarithm to make the base same as the base of the first term completely.

$\implies$ $2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,-\,$ $\log_{2+\sqrt{3}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $\log_{2+\sqrt{3}} {(\sqrt{x^2+1}-x)}$ $\,=\,$ $3$


Simplifying the second logarithmic term

Try rationalising method one more time to simplify the quantity in function form of the second logarithmic term.

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $\log_{2+\sqrt{3}} \Bigg[{(\sqrt{x^2+1}-x) \times \dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}}\Bigg]$ $\,=\,$ $3$

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $\log_{2+\sqrt{3}} \Bigg[{\dfrac{(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)}{\sqrt{x^2+1}+x}}\Bigg]$ $\,=\,$ $3$

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $\log_{2+\sqrt{3}} \Bigg[{\dfrac{{(\sqrt{x^2+1})}^2-x^2}{\sqrt{x^2+1}+x}}\Bigg]$ $\,=\,$ $3$

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $\log_{2+\sqrt{3}} \Bigg[{\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}}\Bigg]$ $\,=\,$ $3$

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $\require{cancel} \log_{2+\sqrt{3}} \Bigg[{\dfrac{\cancel{x^2}+1-\cancel{x^2}}{\sqrt{x^2+1}+x}}\Bigg]$ $\,=\,$ $3$

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $\log_{2+\sqrt{3}} \Bigg[{\dfrac{1}{\sqrt{x^2+1}+x}}\Bigg]$ $\,=\,$ $3$

Use negative exponent rule to write the second logarithmic term in alternative form.

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}^{-1}$ $\,=\,$ $3$

Use power rule of logarithms to write the second logarithmic term in another form.

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,-\,$ $(-1)\log_{2+\sqrt{3}} (\sqrt{x^2+1}+x)$ $\,=\,$ $3$

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^2$ $\,+\,$ $\log_{2+\sqrt{3}} (\sqrt{x^2+1}+x)$ $\,=\,$ $3$


Combining Logarithmic terms

The two logarithmic terms are connected by a plus sign and they can be combined by using the product rule of logarithms to move ahead in solving this logarithmic equation.

$\implies$ $\log_{2+\sqrt{3}} \Big[{({\sqrt{x^2+1}+x})}^2$ $\times$ $(\sqrt{x^2+1}+x)\Big]$ $\,=\,$ $3$

The multiplying factors are same but have different exponents. Hence, try product rule of exponents to merge them.

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^{2+1}$ $\,=\,$ $3$

$\implies$ $\log_{2+\sqrt{3}} {({\sqrt{x^2+1}+x})}^3$ $\,=\,$ $3$


Transformation from Logarithm to exponential form

The logarithmic equation can be transformed in exponential notation by the mutual mathematical relation between them.

$\implies$ ${({\sqrt{x^2+1}+x})}^3$ $\,=\,$ ${(2+\sqrt{3})}^3$


Solving value of x

The exponents of both sides of the equation is same. Therefore, the bases of them should also be equal in value. Now, solve the equation to obtain the required result.

$\implies$ $\sqrt{x^2+1}+x \,=\, 2+\sqrt{3}$

$\implies$ $x-2 \,=\, \sqrt{3}-\sqrt{x^2+1}$

Square both sides and expand to release the equation from root form.

$\implies$ ${(x-2)}^2 \,=\, {(\sqrt{3}-\sqrt{x^2+1})}^2$

$\implies$ $x^2+4-4x \,=\, 3+x^2+1-2\sqrt{3}(\sqrt{x^2+1})$

$\implies$ $\cancel{x^2}+\cancel{4}-4x \,=\, \cancel{x^2}+\cancel{4}-2\sqrt{3}(\sqrt{x^2+1})$

$\implies$ $2\sqrt{3}(\sqrt{x^2+1}) \,=\, 4x$

$\implies$ $\sqrt{3}(\sqrt{x^2+1}) \,=\, \dfrac{4x}{2}$

$\implies$ $\require{cancel} \sqrt{3}(\sqrt{x^2+1}) \,=\, \dfrac{\cancel{4}x}{\cancel{2}}$

$\implies$ $\sqrt{3}(\sqrt{x^2+1}) \,=\, 2x$

Once again, square root both sides to release the equation completely from root form and then proceed to solving the value of $x$.

$\implies$ ${[\sqrt{3}(\sqrt{x^2+1})]}^2 \,=\, {(2x)}^2$

$\implies$ $3(x^2+1) \,=\, 4x^2$

$\implies$ $3x^2+3 \,=\, 4x^2$

$\implies$ $3 \,=\, 4x^2-3x^2$

$\implies$ $3 \,=\, x^2$

$\implies$ $x^2 \,=\, 3$

$\therefore \,\,\,\,\,\,$ $x \,=\, \pm \sqrt{3}$

The value of $x$ are $\sqrt{3}$ and $-\sqrt{3}$ but test them to confirm whether they both are solutions or not.


Validating solutions

Substitute $x = \sqrt{3}$ and evaluate it.

$2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{2-\sqrt{3}} {(\sqrt{x^2+1}-x)}$

$= 2\log_{2+\sqrt{3}} {\Big(\sqrt{{(\sqrt{3})}^2+1}+\sqrt{3}\Big)}$ $\,+\,$ $\log_{2-\sqrt{3}} {\Big(\sqrt{{(\sqrt{3})}^2+1}-\sqrt{3}\Big)}$

$= 2\log_{2+\sqrt{3}} {\Big(\sqrt{3+1}+\sqrt{3}\Big)}$ $\,+\,$ $\log_{2-\sqrt{3}} {\Big(\sqrt{3+1}-\sqrt{3}\Big)}$

$= 2\log_{2+\sqrt{3}} {\Big(\sqrt{4}+\sqrt{3}\Big)}$ $\,+\,$ $\log_{2-\sqrt{3}} {\Big(\sqrt{4}-\sqrt{3}\Big)}$

$= 2\log_{2+\sqrt{3}} (2+\sqrt{3})$ $\,+\,$ $\log_{2-\sqrt{3}} (2-\sqrt{3})$

$= 2 \times 1 \,+\, 1$

$= 2 \,+\, 1$

$= 3$

Therefore, $\sqrt{3}$ is a root of the logarithmic equation.

Now, substitute $x = -\sqrt{3}$ and find the value of the left hand side expression.

$2\log_{2+\sqrt{3}} {(\sqrt{x^2+1}+x)}$ $\,+\,$ $\log_{2-\sqrt{3}} {(\sqrt{x^2+1}-x)}$

$= 2\log_{2+\sqrt{3}} {\Big(\sqrt{{(-\sqrt{3})}^2+1}+(-\sqrt{3})\Big)}$ $\,+\,$ $\log_{2-\sqrt{3}} {\Big(\sqrt{{(-\sqrt{3})}^2+1}-(-\sqrt{3})\Big)}$

$= 2\log_{2+\sqrt{3}} {\Big(\sqrt{3+1}-\sqrt{3}\Big)}$ $\,+\,$ $\log_{2-\sqrt{3}} {\Big(\sqrt{3+1}+\sqrt{3}\Big)}$

$= 2\log_{2+\sqrt{3}} {\Big(\sqrt{4}-\sqrt{3}\Big)}$ $\,+\,$ $\log_{2-\sqrt{3}} {\Big(\sqrt{4}+\sqrt{3}\Big)}$

$= 2\log_{2+\sqrt{3}} {(2-\sqrt{3})}$ $\,+\,$ $\log_{2-\sqrt{3}} {(2+\sqrt{3})}$

Use Rationalizing method to simplify it.

$= 2\log_{2+\sqrt{3}} {\Bigg[(2-\sqrt{3}) \times \dfrac{(2+\sqrt{3})}{(2+\sqrt{3})}\Bigg]}$ $\,+\,$ $\log_{\Bigg[2-\sqrt{3} \large \times \Large \frac{2+\sqrt{3}}{2+\sqrt{3}} \normalsize \Bigg]} {(2+\sqrt{3})}$

$= 2\log_{2+\sqrt{3}} {\Bigg[\dfrac{(2-\sqrt{3})(2+\sqrt{3})}{(2+\sqrt{3})}\Bigg]}$ $\,+\,$ $\log_{\Bigg[\Large \frac{(2-\sqrt{3})(2+\sqrt{3})}{2+\sqrt{3}} \normalsize \Bigg]} {(2+\sqrt{3})}$

$= 2\log_{2+\sqrt{3}} {\Bigg[\dfrac{2^2-{(\sqrt{3})}^2}{(2+\sqrt{3})}\Bigg]}$ $\,+\,$ $\log_{\Bigg[\Large \frac{2^2-{(\sqrt{3})}^2}{2+\sqrt{3}} \normalsize \Bigg]} {(2+\sqrt{3})}$

$= 2\log_{2+\sqrt{3}} {\Bigg[\dfrac{4-3}{(2+\sqrt{3})}\Bigg]}$ $\,+\,$ $\log_{\Bigg[\Large \frac{4-3}{2+\sqrt{3}} \normalsize \Bigg]} {(2+\sqrt{3})}$

$= 2\log_{2+\sqrt{3}} {\Bigg[\dfrac{1}{(2+\sqrt{3})}\Bigg]}$ $\,+\,$ $\log_{\Bigg[\Large \frac{1}{2+\sqrt{3}} \normalsize \Bigg]} {(2+\sqrt{3})}$

$= 2\log_{2+\sqrt{3}} {(2+\sqrt{3})}^{-1}$ $\,+\,$ $\log_{\large {\Big[(2+\sqrt{3})}^{-1}\Big]} {(2+\sqrt{3})}$

$= -2\log_{2+\sqrt{3}} {(2+\sqrt{3})}$ $\,+\,$ $\dfrac{1}{(-1)} \log_{(2+\sqrt{3})} {(2+\sqrt{3})}$

$= -2 \times 1$ $\,+\,$ $(-1) \times 1$

$= -2$ $\,+\,$ $(-1)$

$= -2 \,-\, 1$

$= -3$

It is not the value of the right hand side of the equation. Hence, $-\sqrt{3}$ cannot be the value of the $x$. Therefore, the real value of $x$ is $\sqrt{3}$ and it is the required solution of the logarithmic problem.

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