If f(x) = log2 (x - 2)3 and g(x) = log8 (x - 2), when is (f + g)(x) = 0?
working with logs base 2,
since 8=2^3, g(x) = 1/3 log(x-2)
(f+g)(x) = f(x)+g(x)
= 3log(x-2) + 1/3 log(x-2)
= 10/3 log(x-2)
so, if (f+g)(x) = 0,
log(x-2) = 0
x-2 = 1
x = 3
check
log_2(3-2)^3 = 0
log_8(3-2) = 0
Thanks Steve! I am starting to write these down so that way I will understand them. It is easier for me to go off an example. I appreciate all your help.
To find the value of x when (f + g)(x) = 0, we need to first express (f + g)(x) in terms of f(x) and g(x).
Given f(x) = log2 (x - 2)^3 and g(x) = log8 (x - 2), to find (f + g)(x), we can add the two functions together:
(f + g)(x) = log2 (x - 2)^3 + log8 (x - 2)
To simplify the expression, we can use the logarithmic property of addition:
log2 (x - 2)^3 + log8 (x - 2) = log2 [(x - 2)^3] + log2 (8^(1/3))
Next, we can apply the power rule of logarithms and simplify further:
log2 [(x - 2)^3] + log2 (8^(1/3)) = log2 [(x - 2)^3] + log2 (2)
Since log2 (2) = 1, we can simplify the expression even further:
log2 [(x - 2)^3] + log2 (2) = log2 [(x - 2)^3] + 1
So, (f + g)(x) = log2 [(x - 2)^3] + 1
Now, to find the value of x when (f + g)(x) = 0, we set the expression equal to zero:
log2 [(x - 2)^3] + 1 = 0
Subtracting 1 from both sides, we get:
log2 [(x - 2)^3] = -1
Using the logarithmic property of equality, we can rewrite it as:
2^(-1) = (x - 2)^3
Simplifying, we have:
1/2 = (x - 2)^3
Taking the cube root of both sides, we get:
∛(1/2) = x - 2
To find the value of ∛(1/2), we can calculate:
∛(1/2) ≈ 0.7937
Adding 2 to both sides, we find:
x ≈ 0.7937 + 2 ≈ 2.7937
Therefore, when (f + g)(x) = 0, x is approximately equal to 2.7937.