Question

Log7 (x-6) +log7 x=1

Answer 1

Ответ:

х = 7

Пошаговое объяснение:

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Answer 2

To solve the logarithmic equation log7(x-6) + log7(x) = 1, we can use the properties of logarithms. One property that will be useful here is the product rule of logarithms, which states that log_b(a) + log_b(c) = log_b(a * c).

Using the product rule, we can combine the two logarithms on the left-hand side of the equation:

log7(x-6) + log7(x) = log7((x-6) * x)

Now, the equation becomes:

log7((x-6) * x) = 1

Since the logarithm base is 7, we can rewrite the equation as an exponential equation:

7^1 = (x-6) * x

Now simplify the equation:

7 = x^2 - 6x

To solve this quadratic equation, let's move all terms to one side:

x^2 - 6x - 7 = 0

Now, we need to factor this quadratic equation. It factors as follows:

(x - 7)(x + 1) = 0

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

1. x - 7 = 0 x = 7

2. x + 1 = 0 x = -1

So, the solutions to the original logarithmic equation are x = 7 and x = -1. However, we need to check if both solutions are valid for the original logarithmic equation.

For logarithmic expressions to be valid, their arguments (the values inside the logarithms) must be greater than zero.

Checking x = 7: log7(7 - 6) + log7(7) = log7(1) + log7(7) = 0 + 1 = 1 The solution x = 7 satisfies the equation.

Checking x = -1: log7((-1) - 6) + log7(-1) = log7(-7) + log7(-1) Since both arguments are negative, this is not valid, and we discard this solution.

Therefore, the only valid solution to the equation is x = 7.