CAT 2024 Slot 3 QA Question 12

The equation given is $10^x + \frac{4}{10^x} = \frac{81}{2}$. We find the sum of all distinct real values of $x$ that satisfy this equation. The options are all in terms of logarithms, so I think I'll need to manipulate the equation to solve for $x$.

First, equations involving terms like $a^x$ and $a^{-x}$ can often be simplified by substitution. Maybe I can let $y = 10^x$. That seems like a good idea because it will make the equation easier to handle.

So, substituting $y = 10^x , the equation becomes$y + $\frac{4}{y}$ = $\frac{81}{2}$$. Hmm, this looks like a quadratic equation in terms of$y$. To make it clearer, I can multiply both sides by$y$ to eliminate the denominator. Let's do that:

Multiplying both sides by $y$: $y \cdot y + y \cdot \frac{4}{y} = \frac{81}{2} \cdot y$$Simplifying each term:$$y^2 + 4 = \frac{81}{2} y$$

Now, to make this a standard quadratic equation, I'll bring all terms to one side:

Wait, quadratic equations are usually easier to solve when the coefficients are integers. Maybe I can eliminate the fraction by multiplying the entire equation by 2:

Alright, now I have a quadratic equation in $y$: $2y^2 - 81y + 8 = 0$. I can solve this using the quadratic formula. The quadratic formula is $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where$a = 2 , b = -81 , and $c = 8$.

Compute the discriminant first: Calculating each part: $(-81)^2 = 6561$ $4 \cdot 2 \cdot 8 = 64$So,$$\Delta = 6561 - 64 = 6497$ Hmm, 6497 doesn't look like a perfect square. Let me check if I did the calculation right. 81 squared is indeed 6561, and 428 is 64, so 6561 - 64 is 6497. Okay, so the discriminant is 6497, which is positive, meaning there are two real roots. That makes sense because the original equation is likely to have two solutions for $x$.

Now, applying the quadratic formula:

But $y = 10^x$ must be positive because any exponential function is positive. Both solutions for $y$ are positive because $81 + \sqrt{6497}$ and $81 - \sqrt{6497}$ are both positive.

First, $\sqrt{6497}$ is approximately \$\sqrt{6497}$. Let me estimate \$\sqrt{6497}$. Since 81^2 is 6561, which is 64 more than 6497, so \$\sqrt{6497}$ is approximately 80.6. So, 81 - 80.6 is about 0.4, which is positive. So both roots are positive, which is good because $y = 10^x$ must be positive.

Now, I need to solve for $x$ in each case. Since $y = 10^x , taking the logarithm base 10 of both sides gives:$x = \log_{10} y$ So, for each value of $y , I can find $x$.

Let the two solutions for $y$ as $y_1$ and $y_2$: $y_1 = \frac{81 + \sqrt{6497}}{4}$ $y_2 = \frac{81 - \sqrt{6497}}{4}$ Then, the corresponding $x$ values are: $x_1 = \log_{10} y_1$ $x_2 = \log_{10} y_2$ The question asks for the sum of all distinct real values of $x$. So, I need to compute $x_1 + x_2$.

Using logarithm properties, I know that: So, $$x_1 + x_2 = \log_{10} y_1 + \log_{10} y_2 = \log_{10} (y_1 \cdot y_2)$ Therefore, We find $y_1 \cdot y_2$.

From quadratic equations, the product of the roots $y_1 \cdot y_2 = \frac{c}{a}$. In this case, $a = 2$ and $c = 8 , so:$y_1 \cdot y_2 = $\frac{8}{2}$ = 4x_1 + x_2 = \log_{10} 4$But 4 can be written as$2^2 , so: $\log_{10} 4 = \log_{10} 2^2 = 2 \log_{10} 2$ Therefore, the sum of all distinct real values of $x$ is