Question

Log7(9-x)+1=log7 28 решите уравнение .

Answer 1

ОДЗ: 9-x>0

9>x

x∈(-∞;9)

log7(9-x)+log7(7)=log7(28)

log7(7(9-x))=log7(28)

63-7x=28

7x=35

x=5

Ответ: x=5

Answer 2

To solve the equation, we need to isolate the variable (x) on one side of the equation. Let's start by using the properties of logarithms to simplify the equation:

1. Combine the logarithms on the left side using the properties of logarithms (log a + log b = log (a * b)): log7(9 - x) + 1 = log7(28)

2. Remove the "+1" from the left side of the equation by using the property: log_a(a) = 1 log7(9 - x) = log7(28) - 1

3. Now, we can eliminate the logarithm on both sides using the property: log_a(b) = c is equivalent to a^c = b: 9 - x = 7^(log7(28) - 1)

4. Simplify the right side of the equation: 9 - x = 7^(log7(28) - log7(7)) 9 - x = 7^(log7(28/7)) 9 - x = 7^(log7(4))

5. Since log7(4) is the exponent to which 7 must be raised to get 4, we know that 7^(log7(4)) = 4: 9 - x = 4

6. Now, isolate the variable (x) on one side by subtracting 9 from both sides: x = 9 - 4 x = 5

So, the solution to the equation is x = 5.