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

Question

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

Accepted Answer

Let's solve the equation step by step:

$$
10^x + \frac{4}{10^x} = \frac{81}{2}
$$

**Step 1: Substitute $$ y = 10^x $$ (where $$ y > 0 $$)**

$$
y + \frac{4}{y} = \frac{81}{2}
$$

Multiply both sides by $$ y $$:

$$
y^2 + 4 = \frac{81}{2}y
$$

$$
2y^2 + 8 = 81y
$$

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

**Step 2: Solve the quadratic equation**

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

Using the quadratic formula:

$$
y = \frac{81 \pm \sqrt{81^2 - 4 	imes 2 	imes 8}}{2 	imes 2}
$$
$$
y = \frac{81 \pm \sqrt{6561 - 64}}{4}
$$
$$
y = \frac{81 \pm \sqrt{6497}}{4}
$$

Both roots are positive, so both are valid for $$ y = 10^x $$.

**Step 3: Solve for $$ x $$ using $$ y = 10^x $$**

$$
x_1 = \log_{10}\left(\frac{81 + \sqrt{6497}}{4}ight)
$$
$$
x_2 = \log_{10}\left(\frac{81 - \sqrt{6497}}{4}ight)
$$

But since $$ 81 - \sqrt{6497} $$ is positive (as $$ \sqrt{6497} \approx 80.61 $$), both are valid.

**Step 4: Find the sum of all distinct real values of $$ x $$**

Sum:

$$
x_1 + x_2 = \log_{10}\left(\frac{81 + \sqrt{6497}}{4}ight) + \log_{10}\left(\frac{81 - \sqrt{6497}}{4}ight)
$$

$$
= \log_{10}\left(\frac{(81 + \sqrt{6497})(81 - \sqrt{6497})}{16}ight)
$$

$$
= \log_{10}\left(\frac{81^2 - ( \sqrt{6497} )^2}{16}ight)
$$

$$
= \log_{10}\left(\frac{6561 - 6497}{16}ight)
$$

$$
= \log_{10}\left(\frac{64}{16}ight)
$$

$$
= \log_{10}(4)
$$

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

**Final Answer:**

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

Answer

Answer

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