what is logπ (1/√π)
(log base pi of one over square root of pi)
can you explain how ?? thanks
logπ (1/√π)
= logπ (π^(-1/2) )
= (-1/2) logπ π
= (-1/2)(1)
= -1/2
Certainly! To find log base π of (1/√π), we need to apply the rules of logarithms.
First, let's simplify the expression (1/√π) which is equivalent to (√π)^(-1).
Now, we can rewrite the original expression as logπ((√π)^(-1)).
According to the logarithmic rule, log base b of a^c is equal to c times log base b of a. We can apply this rule to our expression.
Thus, logπ((√π)^(-1)) = -1 * logπ(√π).
Next, let's simplify logπ(√π). Using another logarithmic rule, log base b of √a is equal to (1/2) times log base b of a.
Taking advantage of this rule, logπ(√π) = (1/2) * logπ(π). Since log base b of b is equal to 1, we have logπ(π) = 1.
Now, we can substitute this result back into our expression: logπ(√π) = (1/2) * 1 = 1/2.
Finally, let's substitute this value back into our original expression: -1 * (1/2) = -1/2.
Therefore, logπ (1/√π) is equal to -1/2.