paigecropsey
paigecropsey paigecropsey
  • 10-08-2018
  • Mathematics
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Algebra/Trig. I literally have no clue

AlgebraTrig I literally have no clue class=

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bawbdmo
bawbdmo bawbdmo
  • 10-08-2018

Since the base of our logarithmic function is unknown, we can just call it:

[tex] f(x)=\log_k(x) [/tex]

We are then given the point, (4,-2), so plugging in 4 for x and -2 for f(x) we get:

[tex] -2=\log_{k}(4) \implies \\
-2=\frac{\log{4}}{\log{k}} \implies \\
-2\log{k}=\log{4} \implies \\
-2\log{k}=2\log{2} \implies \\
\log{k}=\frac{2\log{2}}{-2} \implies\\
\log{k}=-\log{2} \implies\\
k=10^{-\log{2}} \implies \\
k=\frac{1}{10^\log{2}}} \implies \\
k=\frac{1}{2} [/tex]

So our base is 1/2.

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