Question

1. log 0.6 (x-5) больше log 0.6 (5x-9) 2. log2 (x-1) меньше 1 + log 2 (3)3. Log 0.7x+log0.7(x-1)=log0.7 2

Answer 1

Ответ:

1. log 0.6 (x-5) > log 0.6 (5x-9)

ОДЗ (область допустимых значений) : x-5>0, 5x-9>0 ⇒ x>5, x>9/5 ⇒ x>5

log 0.6 (x-5) > log 0.6 (5x-9), 0<0.6<1:

x-5<5x-9

9-5<5x-x

4<4x

1<x

Учитываем ОДЗ: x>1, x>5 ⇒ x>5 ⇒ x∈(5; +∞)

2. log2 (x-1) < 1 + log2 (3)

ОДЗ: x-1>0  ⇒  x>1

log2 (x-1) < log2 (2) + log2 (3)

log2 (x-1) < log2 (2*3), 2>1

x-1<2*3

x<6+1

x<7

Учитываем ОДЗ: x<7, x>1 ⇒ x∈(1; 7)

3. log 0.7 (x)+log0.7 (x-1)=log0.7 (2)

ОДЗ: x>0, x-1>0 ⇒ x>0, x>1 ⇒ x>1

log 0.7 (x)+log0.7 (x-1)=log0.7 (2)

log 0.7 (x*(x-1))=log0.7 (2)

x*(x-1)=2

x²-x-2=0

D=(-1)²-4·1·(-2)=1+8=9=3²

x₁=(1-3)/2= -2/2=-1 > 1 ?, то есть условие ОДЗ не выполняется

x₂=(1+3)/2= 4/2 = 2 > 1, то есть условие ОДЗ выполняется

x∈{ 2 }

Answer 2

Прикрепил .............

Answer 3

To solve the given logarithmic inequalities, we need to follow some rules and properties of logarithms. Remember that logarithmic functions have certain properties that can help us simplify and solve these types of inequalities:

1. log(a) + log(b) = log(a * b)
2. log(a) - log(b) = log(a / b)
3. log(a^b) = b * log(a)

Let's solve each inequality step by step:

1. log(0.6, x - 5) > log(0.6, 5x - 9)

To compare the two logarithmic expressions with the same base, we can use the property:

If log(a) > log(b), then a > b.

Using this property, we can rewrite the inequality:

x - 5 > 5x - 9

Now, let's solve for x:

4x > 4 x > 1

So the solution to the first inequality is x > 1.

2. log2(x - 1) < 1 + log2(3)

To solve this inequality, we'll first simplify it using the properties of logarithms:

log2(x - 1) - log2(3) < 1

Now, combine the logarithms:

log2[(x - 1) / 3] < 1

Now, we'll convert the inequality to exponential form:

(x - 1) / 3 < 2^1 (x - 1) / 3 < 2

Next, solve for x:

x - 1 < 6 x < 7

So the solution to the second inequality is x < 7.

3. log(0.7, x) + log(0.7, x - 1) = log(0.7, 2)

To solve this logarithmic equation, we'll use the property:

log(a) + log(b) = log(a * b)

Apply this property to the equation:

log(0.7, x * (x - 1)) = log(0.7, 2)

Since the bases are the same, the arguments must be equal:

x * (x - 1) = 2

Now, we'll solve for x:

x^2 - x = 2 x^2 - x - 2 = 0

Factor the quadratic equation:

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

Now set each factor to zero and solve for x:

x - 2 = 0 or x + 1 = 0 x = 2 or x = -1

However, we must check if the solutions satisfy the original equation:

For x = 2: log(0.7, 2) + log(0.7, 2 - 1) = log(0.7, 2) + log(0.7, 1) = log(0.7, 2) + 0 = log(0.7, 2)

For x = -1: log(0.7, -1) + log(0.7, -1 - 1) = log(0.7, -1) + log(0.7, -2)

Logarithms of negative numbers and zero are not defined in the real number system, so x = -1 is not a valid solution.

Therefore, the only valid solution is x = 2.

So the solution to the third equation is x = 2.