SOLVING LOGARITHMIC EQUATIONS

Note:

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Solve for x in the following equation.

Problem 9.2b:

$17\sin \left( 6x\right) -8=0$

Answers:          There are an infinite number of solutions:

$x=\displaystyle \frac{1}{6} \sin ^{-1}\left( \displaystyle \frac{8}{17}\right) \pm n\left( \displaystyle \frac{\pi }{3}\right)$and $x=\displaystyle \frac{\pi }{6}-\displaystyle \frac{1}{6}\sin ^{-1}\left( \displaystyle \frac{8}{17}\right) \pm \pi \left( \displaystyle \frac{\pi }{3}\right)$ are the exact solutions, and $x\approx 0.08165955\pm n\left( \displaystyle \frac{\pi }{3}\right)$ and $x\approx 0.44193922\pm n\left( \displaystyle \frac{\pi }{3}\right)$ are the approximate solutions.

Solution:

To solve for x, first isolate the sine term.

$$\begin{eqnarray*}&& \\ 17\sin \left( 6x\right) -8 &=&0 \\ && \\ \sin \left( 6x\right) &=&\displaystyle \frac{8}{17} \\ && \\ && \\ && \end{eqnarray*}$$

If we restrict the domain of the cosine function to $-\displaystyle \frac{\pi }{2}\leq 6x\leq \displaystyle \frac{\pi }{2}\rightarrow -\displaystyle \frac{\pi }{12}\leq x\leq \displaystyle \frac{\pi }{12 }$, we can use the arcsin function to solve for x.

$$\begin{eqnarray*}\sin\left( 6x\right) &=&\displaystyle \frac{8}{17} \\ && \\ \... ...ac{8}{17}\right) \\ && \\ x &\approx &0.08165955 \\ && \\ && \end{eqnarray*}$$

The sine of x is positive in the first quadrant and the second quadrant. This means that there are two solutions in the first counterclockwise rotation from 0 to $$\frac{\pi }{3}$$. One angle x terminates in the first quadrant and the second angle terminates in the second quadrant. One solution is $x=\displaystyle \frac{1}{6}\sin ^{-1}\left( \displaystyle \frac{8}{17}\right)$

The period of $\sin x$ is $2\pi$, and the period of $\sin 6x$ is $$\frac{ \pi }{3}.$$ As 6x rotates $\pi$ radians, x rotates $$\frac{\pi }{6}$$Therefore,the second solution is $x=\displaystyle \frac{\pi }{6}-\displaystyle \frac{1}{6}\sin ^{-1}\left( \displaystyle \frac{8}{17}\right) \approx 0.44193922.\bigskip\bigskip$

Since the period is $$\frac{\pi }{3},$$ this means that the values will repeat every $$\frac{\pi }{3}$$ radians. Therefore, the solutions are $x\approx 0.08165955\pm n\left( \displaystyle \frac{\pi }{3}\right)$ and $x=0.44193922\pm n\left( \displaystyle \frac{\pi }{3}\right)$ where n is an integer.

These solutions may or may not be the answers to the original problem. You much check them, either numerically or graphically, with the original equation.

Numerical Check:

Check the answer x=0.08165955

• Left Side: $\qquad 17\sin \left( 6x\right) -8=0\approx 17\sin \left( 6\left( 0.08165955\right) \right) -8=0\approx 0\bigskip$

• Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 0.08165955for x, then 0.08165955 is a solution.

Check the answer x=0.44193922

• Left Side: $\qquad 17\sin \left( 6x\right) -8=0\approx 17\sin \left( 6\left( 0.44193922\right) \right) -8=0\approx 0\bigskip$

• Right Side:        $0\bigskip$

Since the left side equals the right side when you substitute 0.44193922for x, then 0.44193922 is a solution.

Graphical Check:

Graph the equation

$f(x)=17\sin \left( 6x\right) -8.$

Note that the graph crosses the x-axis many times indicating many solutions.

Note the graph crosses at 0.08165955 ( one of the solutions ). Since the period of the function is $$\frac{\pi }{3}\approx 1.04719755$$, the graph crosses again at 0.08165955+1.04719755=1.128857 and again at $0.08165955+2\left( 1.04719755\right) \approx 2.17605465$, etc.

The graph also crosses at 0.44193922 ( another solution we found ). Since the period is $$\frac{\pi }{3}\approx 1.14719755$$, it will crosses again at 0.44193922+1.04719755=1.48913677 and at $0.44193922+2\left( 1.04719755\right) =2.536334$, etc $.\bigskip\bigskip$

If you would like to review the solution to problem 9.2c, click on solution.

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