To solve the equation Log5 (√x-3) = 1, we need to rewrite it in exponential form.
Using the definition of logarithms, we know that Loga(b) = c is equivalent to a^c = b.
So, for our equation Log5 (√x-3) = 1, we have:
5^1 = √x - 3
Simplifying further, we get:
5 = √x - 3
Adding 3 to both sides gives us:
8 = √x
Now, square both sides to get rid of the square root:
8^2 = x
64 = x
Therefore, the solution to the equation Log5 (√x-3) = 1 is x = 64.
To solve the equation Log5 (√x-3) = 1, we need to rewrite it in exponential form.
Using the definition of logarithms, we know that Loga(b) = c is equivalent to a^c = b.
So, for our equation Log5 (√x-3) = 1, we have:
5^1 = √x - 3
Simplifying further, we get:
5 = √x - 3
Adding 3 to both sides gives us:
8 = √x
Now, square both sides to get rid of the square root:
8^2 = x
64 = x
Therefore, the solution to the equation Log5 (√x-3) = 1 is x = 64.