The sum of all distinct real values of $x$ that satisfy the equation $10^x + \frac{4}{10^x} = \frac{81}{2}$, is

This equation looks complicated with both $10^x$ and $\tfrac{1}{10^x}$ terms. Let us use a substitution to simplify it.

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Let $A = 10^x$ where $A > 0$ (since $10^x$ is always positive for any real $x$).

Notice that $\tfrac{4}{10^x} = \tfrac{4}{A}$, so our equation becomes:

$A + \tfrac{4}{A} = \tfrac{81}{2}$

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To eliminate the fraction, we multiply both sides by $A$:

$A^2 + 4 = \tfrac{81A}{2}$

$2A^2 + 8 = 81A$

$2A^2 - 81A + 8 = 0$

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If our quadratic has solutions $A_1$ and $A_2$, then:

$A_1 = 10^{x_1}$ and $A_2 = 10^{x_2}$

As $A = 10^x$

Here is the key insight: We do not need to find the individual values of $A$. We just need $x_1 + x_2$!

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For any quadratic $ax^2 + bx + c = 0$, the product of roots equals $\tfrac{c}{a}$.

In our quadratic $2A^2 - 81A + 8 = 0$:

$a = 2$, $b = -81$, $c = 8$

So: $A_1 \times A_2 = \tfrac{8}{2} = 4$

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Since $A_1 = 10^{x_1}$ and $A_2 = 10^{x_2}$:

$10^{x_1} \times 10^{x_2} = 4$

Using the exponent rule $a^m \times a^n = a^{m+n}$:

$10^{x_1 + x_2} = 4$

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We take logarithm base 10 of both sides:

$\log_{10}(10^{x_1 + x_2}) = \log_{10}(4)$

The logarithm and exponential cancel on the left side:

$x_1 + x_2 = \log_{10}(4)$

Since $4 = 2^2$:

$x_1 + x_2 = \log_{10}(2^2) = 2\log_{10}(2)$

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The sum of all distinct real values of $x$ is $2\log_{10}(2)$.