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 $\displaystyle \frac{\pi }{3}$. 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 $\displaystyle \frac{
\pi }{3}.$ As 6x rotates $\pi $ radians, x rotates $\displaystyle \frac{\pi }{6}$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 $\displaystyle \frac{\pi }{3},$ this means that the values will repeat every $\displaystyle \frac{\pi }{3}$ 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


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


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 $\displaystyle \frac{\pi }{3}\approx
1.04719755$, 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 $\displaystyle \frac{\pi }{3}\approx 1.14719755$, 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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