We can expand the logarithm of a product as a sum of logarithms:

Then using the change of base formula, we can derive the relationship

This immediately tells us that

Notice that none of
can be equal to 1. This is because

for any choice of
. This means we can safely do the following without worrying about division by 0.

so that

Similarly,

so that

So we end up with

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Another way to do this:



Then

So we have

Answer:
You need to have both the opposite and adjacent side of the right angled triangle. Tan = Opp/Adj
Step-by-step explanation:
Tangents are pretty much everywhere. A good example is maybe a slide at the playground. You’d need the tangent to figure out the angle of the slide and how it is.
The radius from the center of a circle to the point of tangency point shows that it would be perpendicular to the tangent line considering that anything with a radius is circular and the tangent line is… a line, making it impossible to be parallel.
Rewrite it from least to greatest and set it up like shown