The product of all solutions of the equation ( e^{(5*(\log_e x)^2) + 3} = x^8 ), ( x > 0 ), is: (1) ( e^{8/5} ) (2) ( e^{6/5} ) (3) ( e^2 ) (4) ( e )

Key Ideas You Need

Logical Chain of Steps

1. Recognise the variable in both exponent and base: $x$ sits inside $\ln$ and also as a base ($x^8$).
2. Remove the outer exponential by applying $\ln$ to both sides. This is legal because $x>0$ (domain of $\ln$).
3. Use substitution $y = \ln x$ so the equation becomes a standard quadratic.
4. Solve the quadratic using the discriminant or factoring.
5. Convert back to $x$ by exponentiating ($e^y$).
6. Multiply the solutions; since both are powers of $e$, simply add their exponents.

These six moves turn a daunting mixed log-exponential problem into a short, syllabus-friendly puzzle.

Imagine a See-Saw of Numbers

1. Think of ln (log base e) as a magical magnifying glass that turns numbers into new ones so we can add or multiply them more easily.

2. Our see-saw equation is

   $e^{5(\ln x)^2 + 3} = x^8$

3. Taking the ln on both sides is like peeking under the same magnifying glass on both ends of the see-saw so the balance is still fair.

4. After this peek, the big scary power falls down to a neat quadratic (something that looks like $ay^2 + by + c = 0$).

5. Quadratics are like easy Lego bricks: solve for $y$ using the formula, then turn back ($x = e^y$) to get our original $x$.

6. Finally, multiply the two $x$ answers together. A fun surprise—you just add their exponents when the bases are both $e$!

That’s it. 😊

Step-by-Step Solution

Given

$e^{5(\ln x)^2 + 3} = x^8, \quad x>0$

1. Take natural log on both sides

   $\ln\!\left(e^{5(\ln x)^2 + 3}\right) = \ln(x^8)$

   $5(\ln x)^2 + 3 = 8\,\ln x$

2. Substitute $y = \ln x$.

   $5y^2 + 3 = 8y$

   Rearrange to standard quadratic:

   $5y^2 - 8y + 3 = 0$

3. Solve the quadratic

   Discriminant:

   $\Delta = (-8)^2 - 4 \cdot 5 \cdot 3 = 64 - 60 = 4$

   Roots:

   $y = \frac{8 \pm \sqrt{4}}{2 \cdot 5} = \frac{8 \pm 2}{10}$

   $y_1 = \frac{10}{10} = 1, \quad y_2 = \frac{6}{10} = \frac{3}{5}$

4. Back-substitute for $x$

   $x_1 = e^{y_1} = e^1 = e$

   $x_2 = e^{y_2} = e^{3/5}$

5. Product of the solutions

   $x_1 \times x_2 = e^1 \cdot e^{3/5} = e^{1 + 3/5} = e^{8/5}$

6. Match with options
   Option (1): $e^{8/5}$

[\boxed{e^{8/5}}]