e)
log6
x log6
x 1 1
x cannot be -3, since log(-3) is undefined
You are correct, I apologize for the mistake.
Upon further analysis, when dealing with logarithmic equations, we need to ensure that the values inside the logarithm are positive.
In this case, for the equation log6(x) + log6(x + 1) = 1, if x = -3, we would have log6(-3) which is undefined. Therefore, x = -3 is not a valid solution.
The only valid solution is x = 2.
To solve the equation log6(x) + log6(x + 1) = 1, we can apply properties of logarithms.
First, we can use the product rule of logarithms, which states that log(ab) = log(a) + log(b). Applying this rule, we can rewrite the equation as:
log6(x(x + 1)) = 1
Next, we can convert the equation into exponential form. In exponential form, loga(b) = c is equivalent to a^c = b. In this case, we can rewrite the equation as:
6^1 = x(x + 1)
Simplifying further, we have:
6 = x^2 + x
Rearranging the equation, we have:
x^2 + x - 6 = 0
Now we can solve this quadratic equation by factoring or using the quadratic formula. Factoring gives us:
(x + 3)(x - 2) = 0
Setting each factor equal to zero, we have two possible solutions:
x + 3 = 0 --> x = -3
x - 2 = 0 --> x = 2
Therefore, the solutions to the equation log6(x) + log6(x + 1) = 1 are x = -3 and x = 2.