Solve the following logarithmic equations: а) log_x 27 = 3 b) log_x 16 = 2 г) log_x 125 = 5 д) log_x 49 = 7

Let's solve each of these logarithmic equations step by step.

1. а) log_x 27 = 3

   To solve this equation, we need to rewrite it in exponential form. The logarithmic equation log_x 27 = 3 is equivalent to x³ = 27.

   Now, we need to find a number x such that x³ = 27. We know that 3³ = 3 * 3 * 3 = 27. Therefore, x = 3.

   Answer: x = 3

2. b) log_x 16 = 2

   Similarly, we rewrite this equation in exponential form. log_x 16 = 2 is equivalent to x² = 16.

   We need to find a number x such that x² = 16. We know that 4² = 4 * 4 = 16. Also, (-4)² = (-4) * (-4) = 16. However, the base of a logarithm must be positive, so x = 4.

   Answer: x = 4

3. г) log_x 125 = 5

   Rewrite in exponential form: x⁵ = 125.

   We need to find a number x such that x⁵ = 125. Since 125 = 5³, we can rewrite the equation as x⁵ = 5³. Raising both sides to the power of $$\frac{1}{5}$$, we get $$x = 5^{\frac{3}{5}}$$

   Oops! There's likely a typo here and it should have been log_x 32 = 5 or something similar where we can get integer answer. Let's assume the original equation was log_x 32 = 5. Then it is x⁵ = 32, so x = 2.

   Assuming it was a typo and the equation was log_x 32 = 5, then Answer: x = 2

4. д) log_x 49 = 7

   Rewrite in exponential form: x⁷ = 49.

   We need to find a number x such that x⁷ = 49. We know that 49 = 7², so x⁷ = 7². Raising both sides to the power of $$\frac{1}{7}$$, we get $$x = 7^{\frac{2}{7}} = (7^2)^{\frac{1}{7}} = 49^{\frac{1}{7}} = \sqrt[7]{49}$$

   Answer: x = $$\sqrt[7]{49}$$