2^x 13^(x+10) = (2^x2) 36^(x2)

The equation is:

$$2^x \times 13^{x+10} = (2^{x^2}) \times 36^{x^2}$$

First, express 36 in terms of its prime factors: $36 = 6^2 = (2 \times 3)^2 = 2^2 \times 3^2$.
Now substitute this into the right side of the equation:

$$2^x \times 13^{x+10} = 2^{x^2} \times (2^2 \times 3^2)^{x^2}$$

Apply the exponent rule $(a^m)^n = a^{mn}$ and $(ab)^n = a^n b^n$:

$$2^x \times 13^{x+10} = 2^{x^2} \times 2^{2x^2} \times 3^{2x^2}$$

Combine the terms with base 2 on the right side using the rule $a^m \times a^n = a^{m+n}$:

$$2^x \times 13^{x+10} = 2^{x^2 + 2x^2} \times 3^{2x^2}$$ $$2^x \times 13^{x+10} = 2^{3x^2} \times 3^{2x^2}$$

To solve for $x$, we take the logarithm (e.g., base 10 logarithm, denoted as $\log$) of both sides. Use the logarithm properties $\log(ab) = \log a + \log b$ and $\log(a^b) = b \log a$:

$$\log(2^x \times 13^{x+10}) = \log(2^{3x^2} \times 3^{2x^2})$$ $$x \log 2 + (x+10) \log 13 = 3x^2 \log 2 + 2x^2 \log 3$$

Expand the terms:

$$x \log 2 + x \log 13 + 10 \log 13 = 3x^2 \log 2 + 2x^2 \log 3$$

Move all terms to one side to form a quadratic equation:

$$(3 \log 2 + 2 \log 3)x^2 - (\log 2 + \log 13)x - 10 \log 13 = 0$$

Use logarithm properties $\log a + \log b = \log(ab)$ and $b \log a = \log(a^b)$:

$$(\log(2^3) + \log(3^2))x^2 - (\log(2 \times 13))x - 10 \log 13 = 0$$ $$(\log 8 + \log 9)x^2 - (\log 26)x - 10 \log 13 = 0$$ $$(\log(8 \times 9))x^2 - (\log 26)x - 10 \log 13 = 0$$ $$(\log 72)x^2 - (\log 26)x - 10 \log 13 = 0$$

This is a quadratic equation where $A = \log 72$, $B = -\log 26$, and $C = -10 \log 13$.

The quadratic formula is $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$.

$$x = \frac{-(-\log 26) \pm \sqrt{(-\log 26)^2 - 4(\log 72)(-10 \log 13)}}{2(\log 72)}$$ $$x = \frac{\log 26 \pm \sqrt{(\log 26)^2 + 40 (\log 72)(\log 13)}}{2 \log 72}$$

This is the exact solution for $x$.

$\log 72 \approx 1.8573$
$\log 26 \approx 1.4150$
$\log 13 \approx 1.1139$

Substitute these values into the quadratic formula:
$A \approx 1.8573$
$B \approx -1.4150$
$C \approx -10 \times 1.1139 = -11.1390$

Calculate the discriminant $\Delta = B^2 - 4AC$:
$\Delta \approx (-1.4150)^2 - 4(1.8573)(-11.1390)$
$\Delta \approx 2.0022 + 82.7285$
$\Delta \approx 84.7307$
$\sqrt{\Delta} \approx \sqrt{84.7307} \approx 9.2050$

Now, calculate the two possible values for $x$:

$$x = \frac{1.4150 \pm 9.2050}{2(1.8573)}$$ $$x = \frac{1.4150 \pm 9.2050}{3.7146}$$ $$x_1 = \frac{1.4150 + 9.2050}{3.7146} = \frac{10.6200}{3.7146} \approx 2.8590$$ $$x_2 = \frac{1.4150 - 9.2050}{3.7146} = \frac{-7.7900}{3.7146} \approx -2.0972$$

The solutions for $x$ are approximately:
$x \approx 2.8590 \text{ or } x \approx -2.0972$