# Thread: Figuring out the number of possible answers in an interval

1. ## Figuring out the number of possible answers in an interval

I was recently exposed to the following problem on my homework and I am stumped:

log(base 5)((sin(x))=-1/2

I have no idea how to proceed. This was my thought process working out the problem:

log(base b)(x) = y

b^y = x

Therefore:

5^(-1/2) = sin x

- sqrt(5) = sin x

.... but I am at a loss to try and derive the number of possible answers from this answer. The interval is [0, 2pi). Thank you.

2. Originally Posted by DaGr8Gatzby
I was recently exposed to the following problem on my homework and I am stumped:

log(base 5)((sin(x))=-1/2

I have no idea how to proceed. This was my thought process working out the problem:

log(base b)(x) = y

b^y = x

Therefore:

5^(-1/2) = sin x

-sqrt(5) = sin x

$\textcolor{red}{5^{-\frac{1}{2}} = \frac{1}{\sqrt{5}} = \sin{x}}$

.... but I am at a loss to try and derive the number of possible answers from this answer. The interval is [0, 2pi). Thank you.
two solutions ... one in quad I and one in quad II

...

3. Ok, say I had to do this without a calculator, what would be the indicator that there are 2 solutions here?

4. Hello, DaGr8Gatzby!

You said you have no idea, but you did quite good!

Solve for $x\!:\;\;\log_5(\sin x)\:=\:\text{-}\tfrac{1}{2}$ .for $0 \leq x < 2\pi$

We have: . $\log_5(\sin x) \:=\:\text{-}\tfrac{1}{2} \quad\Rightarrow\quad \sin x \:=\:5^{-\frac{1}{2}} \:=\:\frac{1}{\sqrt{5}}$

Therefore: . $x \;=\;\arcsin\left(\frac{1}{\sqrt{5}}\right) \;\approx\;\begin{Bmatrix}0.464 \\ 2.678 \end{Bmatrix}$

5. Originally Posted by DaGr8Gatzby
Ok, say I had to do this without a calculator, what would be the indicator that there are 2 solutions here?
the values of sin(x) increase from 0 to 1 when x goes from 0 to pi/2, and decrease from 1 to 0 when x goes from pi/2 to pi.

since $0 < \frac{1}{\sqrt{5}} < 1$ , there are two solutions in the quadrants stated earlier.

6. Thanks guys. This helped me clear things up tremendously. I have other problems and will post more Thank you.