The common logarithm of $98$ plus square root of $x$ squared minus $12$ times $x$ plus $36$ equals to $2$ is the given logarithmic equation. We have to solve this logarithmic equation to find the value of $x$ in the logarithmic problem.

$\log_{10}{\Big[98+\sqrt{x^2-12x+36}\Big]}$ $\,=\,$ $2$

There is only one way for simplifying the given logarithmic equation. It can be done by eliminating the logarithmic form from the logarithmic equation with the mathematical relationship between the exponents and logarithms.

$\implies$ $98+\sqrt{x^2-12x+36}$ $\,=\,$ $10^2$

Now, we can concentrate on simplifying the mathematical equation for solving the variable $x$. Firstly, evaluate the square of ten in the right hand side of the equation. This step allows us to initiate the process of simplifying this equation.

$\implies$ $98+\sqrt{x^2-12x+36}$ $\,=\,$ $100$

$\implies$ $\sqrt{x^2-12x+36}$ $\,=\,$ $100-98$

$\implies$ $\sqrt{x^2-12x+36}$ $\,=\,$ $2$

The left hand expression is in square root form. We cannot move forward unless we eliminate it. Hence, take square both sides of the equation and it eliminates the square root from the left hand side expression of the equation.

$\implies$ $\Big(\sqrt{x^2-12x+36}\Big)^2$ $\,=\,$ $2^2$

$\implies$ $x^2-12x+36$ $\,=\,$ $4$

$\implies$ $x^2-12x+36-4$ $\,=\,$ $0$

$\implies$ $x^2-12x+32$ $\,=\,$ $0$

The given logarithmic equation is simplified as a quadratic equation. Now, we can solve the quadratic equation for finding the value of $x$ in this problem.

$x^2-12x+32$ $\,=\,$ $0$

The middle term of the quadratic expression can be split for factoring it.

$\implies$ $x^2-8x-4x+4 \times 8$ $\,=\,$ $0$

$\implies$ $x(x-8)-4(x-8)$ $\,=\,$ $0$

$\implies$ $(x-8)(x-4) \,=\, 0$

$\implies$ $x-8 = 0\,\,$ and $\,\,x-4 = 0$

$\,\,\,\therefore\,\,\,\,\,\,$ $x = 8\,\,$ and $\,\,x = 4$

Therefore, we have solved that $x$ is equal to $4$ and $x$ is equal to $8$.

In some cases, one value may not satisfy the logarithmic equation. Hence, it is essential to verify the values of a variable by substituting each value in the logarithmic equation.

$\log_{10}{\Big(98+\sqrt{x^2-12+36}\Big)}$ $\,=\,$ $2$

Now, substitute each value of $x$ in the left hand side of the expression to know whether we get the value in the right hand side of the equation.

$\log_{10}{\Big(98+\sqrt{{(4)}^2-12(4)+36}\Big)}$

$=\,\,\,$ $\log_{10}{\Big(98+\sqrt{16-48+36}\Big)}$

$=\,\,\,$ $\log_{10}{\Big(98+\sqrt{4}\Big)}$

$=\,\,\,$ $\log_{10}{\Big(98+2\Big)}$

$=\,\,\,$ $\log_{10}{(100)}$

$=\,\,\,$ $\log_{10}{\Big(10^2\Big)}$

$=\,\,\,$ $2 \times \log_{10}{(10)}$

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

$=\,\,\,$ $2$

The value of the logarithmic expression is equal to two when we substitute $x$ equals to $4$. It is the value in the right hand side expression of the given logarithmic equation in this problem. Therefore, the $x = 4$ is a solution of the given logarithmic equation.

$\log_{10}{\Big(98+\sqrt{{(8)}^2-12(8)+36}\Big)}$

$=\,\,\,$ $\log_{10}{\Big(98+\sqrt{64-96+36}\Big)}$

$=\,\,\,$ $\log_{10}{\Big(98+\sqrt{4}\Big)}$

$=\,\,\,$ $\log_{10}{\Big(98+2\Big)}$

$=\,\,\,$ $\log_{10}{100}$

$=\,\,\,$ $\log_{10}{\Big(10^2\Big)}$

$=\,\,\,$ $2 \times \log_{10}{10}$

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

$=\,\,\,$ $2$

This time also the value of the logarithmic expression is equal to $2$ when we replace $x$ equals to $8$.

The given logarithmic equation is satisfied for both $x \,=\, 4$ and $x \,=\, 8$. Hence, the two values are solutions of the given logarithmic equation.

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