Answer:
[tex]y = log_{\sqrt{(-1)} } (x - 5.5) + 2 - 0.25817i[/tex]
Step-by-step explanation:
The general form of a logarithmic function is given as follows;
[tex]y = log_b (x - h) + k[/tex]
From the given data, we have;
When x = 1, y = -30
Therefore;
[tex]b^{-30 - k} = 1 - h[/tex]
When x = 4, y = 0
Therefore;
[tex]b^{0 - k} = 4 - h[/tex]
[tex]b^{ - k} = 4 - h[/tex]
When x = 5, y = 6
Therefore;
[tex]b^{6 - k} = 5 - h[/tex]
[tex]b^{6 - k} = b^{6} \times b^{- k} = 5 - h[/tex]
[tex]\therefore b^{6} =\dfrac{5 - h}{b^{- k}} = \dfrac{5 - h}{4 - h}[/tex]
When x = 6, y = 4
Therefore;
[tex]b^{4 - k} = 6 - h[/tex]
[tex]b^{4 - k} = b^{4} \times b^{- k} = 6 - h[/tex]
[tex]\therefore b^{4} = \dfrac{6 - h}{b^{- k}} = \dfrac{6 - h}{4 - h}[/tex]
Which gives;
[tex]\therefore \dfrac{b^{6}}{b^{4}} =b^{2} = \dfrac{\dfrac{5 - h}{4 - h}}{\dfrac{6 - h}{4 - h}} = \dfrac{5 - h}{6 - h}[/tex]
When x = 7, y = 2
Therefore;
[tex]b^{2 - k} = 7 - h[/tex]
[tex]b^{2} \times b^{- k} = 7 - h[/tex]
[tex]\therefore b^{2} =\dfrac{7 - h}{4 - h}[/tex]
From which we have;
[tex]\dfrac{5 - h}{6 - h} = \dfrac{7 - h}{4 - h}[/tex]
20 - 5·h - 4·h + h² = 42 - 7·h - 6·h + h²
20 - 9·h = 42 - 13·h
4·h = 22
h = 22/4 = 5.5
[tex]\therefore b^{2} =\dfrac{7 - h}{4 - h} =\dfrac{7 - 5.5}{4 - 5.5} = \dfrac{1.5}{-1.5} = -1[/tex]
b = √(-1)
[tex]b^{ - k} = 4 - h[/tex]
-k = log(-1.5)/log(√-1)
k = 2 - 0.258127·i
The logarithmic function is therefore;
[tex]y = log_{\sqrt{(-1)} } (x - 5.5) + 2 - 0.25817i[/tex]
