86092e…: “Versus the windless bay”

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1 : Compute the following numeric exponential value.

$\log _{\sage {p1b1}}\left ({\sage {p1arg1}}\right ) = \answer {\sage {p1ans1}}$
$\log _{\sage {p1b1}}\left ({\sage {p1arg2}}\right ) = \answer {\sage {p1ans2}}$

A good way to think of logs is to put them in their exponential form. Instead of looking at $\log _{\sage {p1b1}}\left ({\sage {p1arg1}}\right ) = ?$ rewrite it as $\sage {p1b1}^{?} = \sage {p1arg1}$ .

2 : Compute the following numeric exponential value.

$\log _{\sage {p2b1}}\left ({\sage {p2arg1}}\right ) = \answer {\sage {p2ans1}}$
$\log _{\sage {p2b1}}\left ({\sage {p2arg2}}\right ) = \answer {\sage {p2ans2}}$

A good way to think of logs is to put them in their exponential form. Instead of looking at $\log _{\sage {p2b1}}\left ({\sage {p2arg1}}\right ) = ?$ rewrite it as $\sage {p2b1}^{?} = \sage {p2arg1}$ .

3 : Compute the following numeric exponential value.

$\log _{\sage {p3b1}}\left ({\sage {p3arg1}}\right ) = \answer {\sage {p3ans1}}$
$\log _{\sage {p3b1}}\left ({\sage {p3arg2}}\right ) = \answer {\sage {p3ans2}}$

A good way to think of logs is to put them in their exponential form. Instead of looking at $\log _{\sage {p3b1}}\left ({\sage {p3arg1}}\right ) = ?$ rewrite it as $\sage {p3b1}^{?} = \sage {p3arg1}$ .

4 : Compute the following numeric exponential value.

$\log _{\sage {p4b1}}\left ({\sage {p4arg1}}\right ) = \answer {\sage {p4ans1}}$
$\log _{\sage {p4b1}}\left ({\sage {p4arg2}}\right ) = \answer {\sage {p4ans2}}$

A good way to think of logs is to put them in their exponential form. Instead of looking at $\log _{\sage {p4b1}}\left ({\sage {p4arg1}}\right ) = ?$ rewrite it as $\sage {p4b1}^{?} = \sage {p4arg1}$ .