# Find the value of $x$, if $\log_{10} \Big[98$ $+$ $\sqrt{x^2-12x+36}\Big]$ $=$ $2$

The literal $x$ is a variable to represent the quantity $2$ by a quadratic form expression through the common logarithm and the logarithmic equation has be solved to obtain the value of $x$.

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

### Step: 1

Use the fundamental relation between logarithm form and exponential notation and write the logarithmic function in exponential form.

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

Simplify the equation.

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

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

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

### Step: 2

Take square both sides of the equation to bring out the quadratic equation from the square root.

$\implies {\Big[\sqrt{x^2-12x+36}\Big]}^2 = 2^2$

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

Now, once again simply the equation.

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

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

### Step: 3

It is a quadratic equation and solve this equation to find the value of $x$. Here, the factoring method can be used to solve the equation.

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

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

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

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

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

#### Verification

The solution of the logarithmic equation has given two values to $x$ and it is very important to verify both values by substituting them one after one in the given logarithmic expression.

##### Substitute $x = 4$

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $\log_{10} \Big[98+\sqrt{{(4)}^2-12(4)+36}\Big]$

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $\log_{10} \Big[98+\sqrt{16-48+36}\Big]$

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $\log_{10} \Big[98+\sqrt{4}\Big]$

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

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $\log_{10} 100$

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

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

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $2 \times 1$

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

The value of the logarithmic function is $2$ when the variable $x$ is replaced by $4$. Therefore, the value of $x$ is equal to $4$.

##### Substitute $x = 8$

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $\log_{10} \Big[98+\sqrt{{(8)}^2-12(8)+36}\Big]$

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $\log_{10} \Big[98+\sqrt{64-96+36}\Big]$

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $\log_{10} \Big[98+\sqrt{4}\Big]$

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

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $\log_{10} 100$

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

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

$\implies \log_{10} \Big[98+\sqrt{x^2-12x+36}\Big]$ $=$ $2 \times 1$

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

Similarly, the value of the logarithmic function is also $2$ when the variable $x$ is replaced by $8$. Therefore, the value of $x$ is equal to $8$.

The verification process has proved that the values of $x$ are $2$ and $4$.

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