Can someone show how to this question.
Loga(x+3) + Loga(x-3) = 3
I don't know if I'm typing this correctly the a is the base.
So, I'm thinking,
Loga((x+3)(x-3)) = 3
(x+3)(x-3) = 3^a
x^2 - 9 = 3^a,
x^2 = 3^a + 9
x = sqrt(3^a + 9)
but now I'm stuck, I don't know how to proceed. I don't know if this is the answer...
You are mixing up 3^a and a^3
The base of the logarithm (a) is what gwets raised to a power (3, in this case).
loga[(x+3)(x-3)] = 3
loga(x^2-9) = 3
a^3 = x^2-9
x = sqrt(a^3 +9)
Thanks... got it...
To solve the equation Loga(x+3) + Loga(x-3) = 3, you are on the right track.
Starting from Loga((x+3)(x-3)) = 3, you correctly expanded the logarithm using the product rule.
Now, let's continue solving the equation:
(x+3)(x-3) = 3^a
Expanding the left side of the equation, we get:
x^2 - 9 = 3^a
Now, to isolate x, we can add 9 to both sides of the equation:
x^2 = 3^a + 9
To find x, you need to take the square root of both sides of the equation. However, you need to consider that the bases match. Since the base of the logarithm is a, we must use the base a to take the square root. This means we need to express 9 as a power of a:
x^2 = 3^a + a^0 (since a^0 = 1)
Now, we can take the square root of both sides:
√(x^2) = √(3^a + a^0)
Simplifying further:
x = √(3^a + 1)
So, the solution to the equation Loga(x+3) + Loga(x-3) = 3 is x = √(3^a + 1).
Please note that this is an algebraic solution, and it might not be possible to simplify further unless specific values for a or x are given.