Question

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

• A. $3\log_{10} 2$
• B. $\log_{10} 2$
• C. $4\log_{10} 2$
• D. $2\log_{10} 2$

Answer 1

Answer: (D)

$2\log_{10} 2$

Answer 2

Let $y = 10^x$. Then, the equation becomes:
$y + \frac{4}{y} = \frac{81}{2}$
Multiply through by $y$ to eliminate the fraction:
$y^2 + 4 = \frac{81y}{2}$
Multiply through by 2 to clear the denominator:
$2y^2 + 8 = 81y$
Rearrange:
$2y^2 - 81y + 8 = 0$
Now, solve this quadratic equation using the quadratic formula:
$y = \frac{-(-81) \pm \sqrt{(-81)^2 - 4(2)(8)}}{2(2)}$
$y = \frac{81 \pm \sqrt{6561 - 64}}{4} = \frac{81 \pm \sqrt{6497}}{4}$
Taking the roots, we find that $y = 10^x$, so:
$\boxed{2\log_{10}2}$
Therefore, the sum of all distinct real values of $x$ is $2\log_{10}2$.