Question

2 log 4(2x+3)=1+log4x+log4(2x-1)

Answer 1

$2log_4(2x+3)=1+log_4x+log_4(2x-1)\\\\ODZ:\; \; \left \{ {{2x+3>0\; ,\; \; x>0} \atop {2x-1>0}} \right. \; \; \left \{ {{x>0} \atop {x>0,5}} \right. \; \; \Rightarrow \; \; x>0,5\\\\log_4(2x+3)^2=log_4(4x\cdot (2x-1))\\\\4x^2+12x+9=8x^2-4x\\\\4x^2-16x-9=0\; ,\; \; D/4=100\; ,\; \; x_{1,2}=\frac{8\pm 10}{4}\\\\x_1=-0,5\notin ODZ\; ,\; \; x_2=4,5\\\\Otvet:\; \; x=4,5\; .$

Answer 2

Решение во вложении.
Ответ: 4,5

Answer 3

To solve the equation 2log₄(2x+3) = 1 + log₄x + log₄(2x-1), we'll use logarithmic properties to simplify the equation and find the value of x.

Step 1: Apply logarithmic properties

• The logarithmic property logₐ(b) + logₐ(c) = logₐ(b * c) can be used to combine the logarithms on the right-hand side of the equation.

2log₄(2x+3) = log₄(x) + log₄(2x-1) + 1

• The logarithmic property logₐ(b * c) = logₐ(b) + logₐ(c) allows us to rewrite the left-hand side of the equation.

log₄((2x+3)²) = log₄(x(2x-1)) + 1

Step 2: Eliminate the base 4 logarithm

• If two logarithms are equal, then the contents inside the logarithms must be equal as well.

(2x+3)² = x(2x-1) * 4

Step 3: Solve for x

• Expand and simplify the equation.

4x² + 12x + 9 = 4x² - x

Step 4: Bring all terms to one side of the equation.

0 = 4x² - 4x² + x + 9 - 12x

0 = x - 3

Step 5: Solve for x.

x = 3

Thus, the solution to the equation is x = 3. It is essential to check that this solution satisfies the original equation since taking the logarithm of non-positive values is not valid in real numbers. If x = 3 does not satisfy the original equation, then it is an extraneous solution, and there may be no valid solutions.