The sum of all distinct real values of x that satisfy the equation:

10^x + 4 /10^x = 81/2

is:

The correct option is 2log₁₀ 2.

To find the sum of all distinct real values of $x$ satisfying the given equation, we start with the equation itself:

${10}^x+\frac{4}{{10}^x}=8\frac{1}{2}$

Note that the fraction on the right-hand side, $8\frac{1}{2}$ (eight and a half), can be written as the improper fraction $\frac{17}{2}$.
Thus, the equation is:

${10}^x+\frac{4}{{10}^x}=\frac{17}{2}$

To simplify, let us substitute $y={10}^x$ (where $y>0$ since $x$ is a real number). Substituting this into the equation gives:

$y+\frac{4}{y}=\frac{17}{2}$

To clear the denominators, we multiply the entire equation by $2y$:

$2y(y+\frac{4}{y})=2y\left(\frac{17}{2}\right)$

This simplifies to a quadratic equation in terms of $y$:

$2y^2+8=17y$

Rearranging the terms into standard form:

$2y^2-17y+8=0$

We can solve this quadratic equation by splitting the middle term:

$2y^2-16y-y+8=0$

Factoring by grouping:

$2y(y-8)-1(y-8)=0$

Which yields:

$(2y-1)(y-8)=0$

Thus, the two solutions for $y$ are:

$y=\frac{1}{2}\text{or}y=8$

Now, we substitute $y={10}^x$ back to solve for $x$:

Case 1: ${10}^x=\frac{1}{2}$
Taking the base-10 logarithm on both sides:

$x={\log }_{10}\left(\frac{1}{2}\right)=-{\log }_{10}2$

Case 2: ${10}^x=8$
Taking the base-10 logarithm on both sides:

$x={\log }_{10}8={\log }_{10}\left(2^3\right)=3{\log }_{10}2$

To find the sum of all distinct real values of $x$, we add these two values together:

$\text{Sum}=-{\log }_{10}2+3{\log }_{10}2$

Simplifying this sum gives:

$\text{Sum}=2{\log }_{10}2$