SOLVING LOGARITHMIC EQUATIONS


Note:

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Solve for x in the following equation.


Problem 9.2a:

$8\sin\left( 4x\right) -3=0$



Answers:         There are an infinite number of solutions:

$x=\displaystyle \frac{1}{4} \sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \pm n\left( \displaystyle \frac{\pi }{2}\right) $and $x=\displaystyle \frac{\pi }{4}-\displaystyle \frac{1}{4}\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \pm \pi \left( \displaystyle \frac{\pi }{2}\right) $ are the exact solutions, and $x\approx 0.096099\pm n\left( \displaystyle \frac{\pi }{2}\right) $ and $x\approx 0.689299\pm n\left( \displaystyle \frac{\pi }{2}\right) $ are the approximate solutions.




Solution:


To solve for x, first isolate the sine term.

\begin{eqnarray*}&& \\ 8\sin\left( 4x\right) -3 &=&0 \\ && \\ \sin\left( 4x\right) &=&\displaystyle \frac{3}{8} \\ && \\ && \\ && \end{eqnarray*}


If we restrict the domain of the cosine function to $-\displaystyle \frac{\pi }{2}\leq 4x\leq \displaystyle \frac{\pi }{2}\rightarrow -\displaystyle \frac{\pi }{8}\leq x\leq \displaystyle \frac{\pi }{8}$, we can use the arcsin function to solve for x.

\begin{eqnarray*}\sin\left( 4x\right) &=&\displaystyle \frac{3}{8} \\ && \\ \s... ...ht) &=&\sin ^{-1}\left( \displaystyle \frac{3}{8} \right) \\ && \end{eqnarray*}
\begin{eqnarray*}&&\\ 4x &=&\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \... ...ac{3}{8}\right) \\ && \\ x &\approx &0.096099193 \\ && \\ && \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 $2\pi $. One angle 4x terminates in the first quadrant and the second angle terminates in the second quadrant. One solution is $x=\displaystyle \frac{1}{4}\sin ^{-1}\left( \displaystyle \frac{3}{8}\right) $


The period of $\sin x$ is $2\pi $, and the period of $\sin 4x$ is $\displaystyle \frac{ \pi }{2}.$ Therefore,the second solution is $x=\displaystyle \frac{\pi }{4}-\displaystyle \frac{1}{4} \sin ^{-1}\left( \displaystyle \frac{3}{8}\right) \approx 0.689299.\bigskip\bigskip $

Since the period is $\displaystyle \frac{\pi }{2},$ this means that the values will repeat every $\displaystyle \frac{\pi }{2}$ radians. Therefore, the solutions are $ x\approx 0.096099193\pm n\left( \displaystyle \frac{\pi }{2}\right) $ and $x=\pm n\left( \displaystyle \frac{\pi }{2}\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=0.096099193


Since the left side equals the right side when you substitute 0.096099193for x, then 0.096099193 is a solution.




Check the answer x=0.689299


Since the left side equals the right side when you substitute 0.689299 for x, then 0.689299 is a solution.




Graphical Check:


Graph the equation

$f(x)=8\sin\left( 4x\right) -3.$

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


Note the graph crosses at 0.096099193 ( one of the solutions ). Since the period of the function is $\displaystyle \frac{\pi }{2}\approx 1.5707963$, the graph crosses again at 0.096099193+1.5707963=1.6668955 and again at $0.096099193+2\left( 1.5707963\right) \approx 3.237691793$, etc.

The graph also crosses at 0.689299 ( another solution we found ). Since the period is $\displaystyle \frac{\pi }{2}\approx 1.5707963$, it will crosses again at 0.689299+1.5707963=2.2600953 and at $0.689299+2\left( 1.5707963\right) =3.83089$, etc $.\bigskip\bigskip $

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


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