Question

Log3√x-5+log3(2x-3)½= 1 решите уравнение

Answer 1

ОДЗ x-5>0; x>5
        2x-3>0;x>1.5
Общее одз x>5

√(x-5)*√(2x-3)=3
(x-5)(2x-3)=9
2x^2-13x+6=0
D=169-48=121
x1=(13+11)/4=6; x2=(13-11)/4=0.5-не подходит по одз
Ответ х=6

Answer 2

To solve the equation, we'll follow these steps:

1. Combine the logarithmic terms using logarithmic rules.
2. Isolate the radical term (if any).
3. Solve for x.

The equation is:

log₃√(x - 5) + log₃√(2x - 3) = 1

Step 1: Combine the logarithmic terms using logarithmic rules. In this case, we can use the logarithm property: logₐ(b) + logₐ(c) = logₐ(b * c)

log₃√[(x - 5) * (2x - 3)^(1/2)] = 1

Step 2: Isolate the radical term.

3^1 = √[(x - 5) * (2x - 3)^(1/2)]

Now, square both sides to get rid of the radical:

3^2 = [(x - 5) * (2x - 3)^(1/2)]^2

9 = (x - 5) * (2x - 3)

Step 3: Solve for x.

Expand the right side of the equation:

9 = 2x^2 - 13x + 15

Move all terms to the left side:

2x^2 - 13x + 15 - 9 = 0

Simplify:

2x^2 - 13x + 6 = 0

Now, we can solve this quadratic equation using factoring or the quadratic formula. Let's use factoring:

2x^2 - 13x + 6 = 0

Factor the quadratic:

(2x - 1)(x - 6) = 0

Now, set each factor to zero and solve for x:

1. 2x - 1 = 0 2x = 1 x = 1/2

2. x - 6 = 0 x = 6

So, the solutions to the equation are x = 1/2 and x = 6.