Question

Log2 (2x+1)=2log2 3-log2 (x-4)

Answer 1

$log_2(2x+1)=2log_23-log_2(x-4);$
ОДЗ:  $\left \{ {{2x+1>0} \atop {x-4>0}} \right. \Rightarrow \left \{ {{x>-\frac{1}{2} } \atop {x>4}} \right.$ $\Rightarrow x>4$
$log_2(2x+1)+log_2(x-4)=log_29;$
$log_2(2x+1)(x-4)=log_29;$
$(2x+1)(x-4)=9;$
$(2x+1)(x-4)=9;2 x^{2}-7x-13=0;D=49+104=153;$
$x_1= \frac{7- \sqrt{153}}{4};$ посторонний корень
$x_2= \frac{7+ \sqrt{153}}{4};$
Ответ:  $x= \frac{7+ \sqrt{153}}{4}$

Answer 2

To solve the equation log2(2x + 1) = 2log2(3) - log2(x - 4), you can use logarithmic properties to simplify it and isolate x. Here are the steps to solve the equation:

Step 1: Apply the properties of logarithms to simplify the equation.

log2(2x + 1) = log2(3^2) - log2(x - 4)

Step 2: Use the properties of logarithms to combine the terms on the right side of the equation.

log2(2x + 1) = log2(9) - log2(x - 4)

Step 3: Use the properties of logarithms to simplify the right side of the equation.

log2(2x + 1) = log2(9/(x - 4))

Step 4: Since the bases of the logarithms are the same (base 2), you can set the arguments equal to each other.

2x + 1 = 9/(x - 4)

Step 5: Solve for x.

2x(x - 4) + 1(x - 4) = 9

2x^2 - 8x + x - 4 = 9

2x^2 - 7x - 4 - 9 = 0

2x^2 - 7x - 13 = 0

Now, you have a quadratic equation. You can use the quadratic formula to solve for x:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 2, b = -7, and c = -13. Plug these values into the quadratic formula:

x = (7 ± √((-7)^2 - 4 * 2 * (-13))) / (2 * 2)

x = (7 ± √(49 + 104)) / 4

x = (7 ± √153) / 4

So, the two possible solutions for x are:

x1 = (7 + √153) / 4 x2 = (7 - √153) / 4