Unit Circle and Logarithm Problem #1 (Trigonometry)

Unit Circle and Logarithm #1 Question:

The point $\left(log_b\left(\sqrt{b}\right),\:y\right)$ where $b>0,\:b=1$, is on the terminal arm of angle $\theta$ drawn in the standard position of the unit circle. Find the angle $\theta$ between 0 and $\pi$.

Answer:

For unit circle and logarithm #1 problem, we need to apply our knowledge on logarithm rules and on the unit circle.

First, let’s simplify the x-axis value of $\left(log_b\left(\sqrt{b}\right),\:y\right)$.

$$log_b\left(\sqrt{b}\right)$$

We can express $\sqrt{b}$ as $b^{\frac{1}{2}}$.

Then, we can apply the Log of Exponent Rule to simplify the x-coordinate value.

$$log_b\left(b^n\right)=n$$

$$log_b\left(b^{\frac{1}{2}}\right)=\frac{1}{2}$$

Since the x-coordinate can be expressed as $\frac{1}{2}$, then we need to use the unit circle to find the angle $\theta$ where the point $\left(\frac{1}{2},\:y\right)$ lies. As stated on the question, the angle $\theta$ is between 0 and $\pi$ (Quadrant 1).

A unit circle is posted below.

Unit Circle and Logarithm #1

Accordingly, only angle $60^{\circ }$ or $\frac{\pi }{3}$ has an ordered pair $\left(\frac{1}{2},\:\frac{\sqrt{3}}{2}\right)$ in Quadrant 1. And, we would be using radians as given on the interval.

Besides $\frac{\pi }{3}$, no other angles in Quadrant 1 have $x=\frac{1}{2}$.

Thus, angle $\theta =\frac{\pi }{3}$.

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