Solve this equation
log8 (x2 - 7x) = 1
To solve the equation log8 (x^2 - 7x) = 1, we can use the properties of logarithms.
The property we will use is that if logA (B) = C, then A^C = B.
Using this property, we can rewrite the equation as:
8^1 = x^2 - 7x
Simplifying further, we have:
8 = x^2 - 7x
Now, we have a quadratic equation. To solve for x, we should rearrange the equation in standard quadratic form, which is ax^2 + bx + c = 0.
So, let's move all terms to one side to get:
x^2 - 7x - 8 = 0
Now, we can solve this quadratic equation by factoring or using the quadratic formula.
To factor the equation, we need to find two numbers whose sum is -7 (coefficient of x) and whose product is -8 (constant term). After some calculation, we find that -8 = -1 * 8 and -1 + 8 = -7. Hence, the factored form of the equation is:
(x - 8)(x + 1) = 0
Now, we can set each factor equal to zero and solve for x:
x - 8 = 0 or x + 1 = 0
Solving each equation separately:
For x - 8 = 0, we add 8 to both sides:
x = 8
For x + 1 = 0, we subtract 1 from both sides:
x = -1
Therefore, the solutions to the equation log8 (x^2 - 7x) = 1 are x = 8 and x = -1.