You have provided a set of equations to solve for $x$. I will go through each equation and explain how to solve it:

1. $169^x = 13^{2x}$

   • Recognize that $169 = 13^2$.
   • Therefore, $(13^2)^x = 13^{2x}$.
   • Simplifying, $13^{2x} = 13^{2x}$.
   • Since the bases and exponents are equal, any value for $x$ satisfies the equation.

2. $7^x = \frac{1}{49}$

   • Recognize that $49 = 7^2$, so $\frac{1}{49} = 7^{-2}$.
   • Therefore, $7^x = 7^{-2}$.
   • Solving for $x$, $x = -2$.

3. $\left(\frac{3}{5}\right)^{3x+1} = \frac{25}{9}$

   • Rewrite $\frac{25}{9}$ as $\left(\frac{5}{3}\right)^2$.
   • Therefore, $\left(\frac{3}{5}\right)^{3x+1} = \left(\frac{5}{3}\right)^2$.
   • Which is $\left(\frac{3}{5}\right)^{3x+1} = \left(\frac{3}{5}\right)^{-2}$.
   • Equating the exponents, $3x+1 = -2$.
   • Solving for $x$, $x = -1$.

4. $4^{3x+2} < 64$

   • Recognize $64 = 4^3$.
   • Therefore, $4^{3x+2} < 4^3$.
   • Equating bases, $3x+2 < 3$.
   • Solving for $x$, $x < \frac{1}{3}$.

5. $\left(\frac{9}{27}\right)^{6x-1} \geq \left(\frac{27}{9}\right)^{-x+6}$

   • Simplify to $\left(\frac{1}{3}\right)^{6x-1} \geq \left(3\right)^{-x+6}$.
   • Which is $3^{-6x+1} \geq 3^{-x+6}$.
   • Equating exponents, $-6x+1 \geq -x+6$.
   • Solving for $x$, $x \leq -1$.

6. $4^{x+1} = \frac{1}{64}$

   • Recognize $64 = 4^3$, so $\frac{1}{64} = 4^{-3}$.
   • Therefore, $4^{x+1} = 4^{-3}$.
   • Solving for $x$, $x = -4$.

7. $5^{3x+8} = 25^x$

   • Rewrite $25 = 5^2$.
   • Therefore, $5^{3x+8} = 5^{2x}$.
   • Equating the exponents, $3x+8 = 2x$.
   • Solving for $x$, $x = -8$.

8. $4^{5x-13} = \frac{1}{8^x}$

   • Rewrite $8 = 2^3$, so $\frac{1}{8^x} = 2^{-3x}$, and $2 = 4^{1/2}$.
   • Therefore, $4^{5x-13} = 4^{-1.5x}$.
   • Equating exponents, $5x-13 = -1.5x$.
   • Solving for $x$, $x = \frac{13}{6.5} = 2$.

9. $10^{2x} > 100^{-2x-5}$

   • Rewrite $100 = 10^2$.
   • Therefore, $10^{2x} > 10^{-4x-10}$.
   • Equating bases, $2x > -4x-10$.
   • Solving for $x$, $x > -2$.

10. $49^x = 343^{2x-3}$

• Recognize $49 = 7^2$ and $343 = 7^3$.
• Therefore, $(7^2)^x = (7^3)^{2x-3}$.
• Simplifying, $7^{2x} = 7^{6x-9}$.
• Equating exponents, $2x = 6x-9$.
• Solving for $x$, $x = \frac{9}{4}$.

Would you like details or have any questions about these solutions?

To further expand your understanding, consider the following questions:

1. What would be the effect on the equation $4^{3x+2} < 64$ if the "<" were replaced with "≥"?
2. How do the properties of logarithms help in solving these types of equations?
3. What is the general method for solving exponential inequalities, like in question 4?
4. Can you think of a real-life situation where exponential equations might be used?
5. How would you graph one of these exponential equations to visually verify the solution?

Tip: Always double-check your work by substituting the solution back into the original equation to verify that it indeed satisfies the conditions set by the equation.