Question

Log4(x-6)+logx-6(16)=3

Answer 1

Решение во вложении

Answer 2

To solve the equation log4(x - 6) + log(x - 6) = 16 = 3, we can use logarithmic properties to combine the logarithms. The following steps will lead us to the solution:

Step 1: Combine the logarithms using the product rule of logarithms: log4(x - 6)(x - 6) = 3

Step 2: Simplify the expression inside the logarithm: log4((x - 6)^2) = 3

Step 3: Use the definition of logarithms to rewrite the equation in exponential form: 4^3 = (x - 6)^2

Step 4: Calculate 4^3: 64 = (x - 6)^2

Step 5: Take the square root of both sides: ±√64 = √((x - 6)^2)

Step 6: Solve for x: x - 6 = ±√64

Step 7: Evaluate the square root of 64: x - 6 = ±8

Step 8: Solve for x in both cases: Case 1: x - 6 = 8 x = 8 + 6 x = 14

Case 2: x - 6 = -8 x = -8 + 6 x = -2

So, the two solutions to the equation are x = 14 and x = -2. However, we need to check if these values are valid since the logarithm of a negative number is not defined in the real number system.

For x = 14: log4(14 - 6) + log(14 - 6) = log4(8) + log(8) = 3 + 0 = 3 (valid)

For x = -2: log4(-2 - 6) + log(-2 - 6) = log4(-8) + log(-8)

Since the logarithm of a negative number is not defined in the real number system, x = -2 is not a valid solution.

Thus, the only valid solution to the equation is x = 14.