Question

Log2(3x-1)=5-log2(x+1)

Answer 1

$log_2(3x-1)=5-log_2(x+1)$

ОДЗ:

$\left \{ {{3x-1>0} \atop {x+1>0}} \right.$

$\left \{ {{3x>1} \atop {x>-1}} \right.$

$\left \{ {{x>\frac{1}{3}} \atop {x>-1}} \right.$

$x$ ∈ $(\frac{1}{3} ; +$ ∞ $)$

$log_2(3x-1)+log_2(x+1)=5$

$log_2[(3x-1)*(x+1)]=log_22^5$

$log_2(3x^2+2x-1)=log_232$

$3x^2+2x-1=32$

$3x^2+2x-1-32=0$

$3x^2+2x-33=0$

$D=2^2-4*3*(-33)=4+396=400=20^2$

$x_1=\frac{-2+20}{6} =3$

$x_2=\frac{-2-20}{6} =-\frac{11}{3} =-3\frac{2}{3}$ ∉ ОДЗ

Ответ: $3$

Answer 2

To solve the equation Log2(3x-1) = 5 - Log2(x+1), we need to use some logarithmic properties to simplify and solve for x.

Step 1: Combine the logarithms on the right side using the logarithmic property log_a(b) - log_a(c) = log_a(b/c):

Log2(3x-1) = 5 - Log2(x+1)

Step 2: Move the logarithm on the left side to the right side:

Log2(3x-1) + Log2(x+1) = 5

Step 3: Combine the logarithms on the right side using the logarithmic property log_a(b) + log_a(c) = log_a(b * c):

Log2[(3x-1)(x+1)] = 5

Step 4: Rewrite the equation in exponential form. In general, log_a(b) = c is equivalent to a^c = b:

2^5 = (3x-1)(x+1)

Step 5: Simplify the equation:

32 = 3x^2 + 2x - 1

Step 6: Move all terms to one side of the equation to set it equal to zero:

3x^2 + 2x - 1 - 32 = 0

Step 7: Combine like terms:

3x^2 + 2x - 33 = 0

Now, we have a quadratic equation in standard form (ax^2 + bx + c = 0). To solve for x, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 3, b = 2, and c = -33.

Step 8: Plug the values into the quadratic formula:

x = (-(2) ± √((2)^2 - 4 * 3 * (-33))) / 2 * 3

x = (-2 ± √(4 + 396)) / 6

x = (-2 ± √400) / 6

Step 9: Evaluate the square root:

x = (-2 ± 20) / 6

Now, we have two solutions:

1. x = (20 - 2) / 6 = 18 / 6 = 3
2. x = (-20 - 2) / 6 = -22 / 6 = -11/3

So, the solutions to the equation are x = 3 and x = -11/3.