Question

1)log₃(x+2)=3-log₃(5x+4) 2)log₃(x+1)+log₃(x+3)=1 3)log₂(-x²-2x+3) = log₂(x²+x-2)

Answer 1

Решение задания смотри на фотографии

Answer 2

To solve these logarithmic equations, we'll apply logarithmic rules and algebraic manipulations. Let's tackle each equation one by one:

1. log₃(x+2) = 3 - log₃(5x+4)

Step 1: Combine logarithms using the quotient rule of logarithms. log₃(x+2) + log₃(5x+4) = 3

Step 2: Apply the product rule of logarithms to condense the left side. log₃[(x+2)(5x+4)] = 3

Step 3: Rewrite the equation in exponential form. 3^3 = (x+2)(5x+4)

Step 4: Simplify and solve for x. 27 = 5x² + 14x + 8 5x² + 14x - 19 = 0

Using the quadratic formula or factoring, you can find the solutions for x.

2. log₃(x+1) + log₃(x+3) = 1

Step 1: Combine logarithms using the product rule of logarithms. log₃[(x+1)(x+3)] = 1

Step 2: Rewrite the equation in exponential form. 3^1 = (x+1)(x+3)

Step 3: Simplify and solve for x. 3 = x² + 4x + 3 x² + 4x - 0 = 0

Factor the equation and solve for x.

3. log₂(-x²-2x+3) = log₂(x²+x-2)

Step 1: Set the arguments inside the logarithms equal to each other. -x² - 2x + 3 = x² + x - 2

Step 2: Move all terms to one side of the equation. 2x² + 3x - 5 = 0

Step 3: Solve the quadratic equation for x.

It's important to note that in the third equation, we have a logarithm with a negative argument. Logarithms are defined only for positive arguments, so this equation does not have real solutions. Thus, the equation has no valid solutions.

Make sure to double-check your solutions and discard any extraneous solutions that may arise from the logarithmic manipulations.