The image contains a problem about solving exponential equations. Let me transcribe and solve the equations from the image.

Problem Set:

Задание 2. Решить показательное уравнение:

1. Вариант 1:
   $3^{x-2} \cdot 3^x = 27$

2. Вариант 2:
   $4^{x-1} = 16$

3. Вариант 3:
   $9 \cdot 3^x + 3^{x+2} = 36$

4. Вариант 4:
   $3^x + 3^{x+3} = 81$

Solutions:

1. For Variant 1:
   $3^{x-2} \cdot 3^x = 27$
   Combine powers using the property $a^m \cdot a^n = a^{m+n}$:
   $3^{x-2 + x} = 27$
   $3^{2x-2} = 27$
   Rewrite 27 as $3^3$:
   $3^{2x-2} = 3^3$
   Equate the exponents:
   $2x - 2 = 3$
   $2x = 5$
   $x = \frac{5}{2}$.

2. For Variant 2:
   $4^{x-1} = 16$
   Rewrite 16 as $4^2$:
   $4^{x-1} = 4^2$
   Equate the exponents:
   $x - 1 = 2$
   $x = 3$.

3. For Variant 3:
   $9 \cdot 3^x + 3^{x+2} = 36$
   Rewrite $9$ as $3^2$:
   $3^2 \cdot 3^x + 3^{x+2} = 36$
   Combine powers:
   $3^{x+2} + 3^{x+2} = 36$
   Factorize:
   $2 \cdot 3^{x+2} = 36$
   Divide by 2:
   $3^{x+2} = 18$
   Rewrite $3^{x+2}$ as $3^x \cdot 3^2 = 18$:
   $9 \cdot 3^x = 18$
   Divide by 9:
   $3^x = 2$.
   The solution is $x = \log_3(2)$.

4. For Variant 4:
   $3^x + 3^{x+3} = 81$
   Rewrite $3^{x+3}$ as $3^x \cdot 3^3$:
   $3^x + 27 \cdot 3^x = 81$
   Combine terms:
   $28 \cdot 3^x = 81$
   Divide by 28:
   $3^x = \frac{81}{28}$.
   The solution is $x = \log_3\left(\frac{81}{28}\right)$.

Would you like me to explain any specific step in detail or provide alternative approaches?

Follow-up Questions:

1. What are the properties of exponents used in solving these equations?
2. How do you rewrite numbers like 27 or 16 as powers of their bases?
3. Can logarithms be used to solve any of these problems more directly?
4. How would the solution change if the base of the exponentials were different?
5. How can you check the solutions numerically or graphically?

Tip:

When solving exponential equations, always try to express all terms with the same base to simplify the equations systematically.