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Math 220

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Writing in Mathematics Exercises
119. Explain how to solve an exponential equation when both sides can be written as a power of the same base. a. An exponential equation is defined as an equation that contains a variable in an exponent. In order to solve an exponential equation we need to look at the steps that are required. Exponential equations that have the same base are in the form of If bm=bn. When we see an equation of exponents with the same base we will find the answer by setting the exponents equal to each other. The formula that we can look at to understand how to solve an exponential equations is defined as If bm=bn, then m=n.
The steps that we will take are as follows: 1. Rewrite the equation in the form bm=bn. 2. Set m=n. 3. Solve for the variable.
120. Explain how to solve an exponential equation when both sides cannot be written as a power of the same base. Use 3x=140 in your explanation. a. In order to solve an exponential equation when both sides are not written with the same base, we need to use logarithms. To convert an exponential equation into logarithmic form we look at the formulas below: by=x is equivalent to y=logbx Using 3x=140, we would solve the problem by performing the following steps: 1. Isolate the exponential equation 2. Take the natural logarithm on both sides of the equation for bases other than 10. Take the common logarithm on both sides of the equation for base 10. 3. Simplify using one of the following properties: a. lnbx=x or lnex=x or log10x=x 4. Solve for the variable.

3x=140 ln3x=ln140 xln3=ln140 x= ln140ln3 x≈4.49 (using a calculator)

121. Explain the difference between solving log3x-1=4 and log3x-1= log34. The difference between solving the two equations above is that the second equation can be solved by setting (x-1) = 4 because the logarithmic base is the same. The first equation above needs to be solved by converting the equation into an exponential equation. log3x-1=4 log3x-1= log34
34=x-1 x-1=4
81=x-1 x=5

122. I believe that the 17% risk as a cutoff percentage is a reasonable percentage risk. If I were to change the rate, I would want the percentage risk to be at 10%, but I think the amount of alcohol would be very low. The equation is performed below:
10=6e12.77x e12.77x= 106 lne12.77x=ln⁡(106) 12.77x=ln⁡(106) x=ln⁡(106)12.77 x≈0.04
This blood alcohol level would be very low, which would be much safer, but I think that more people would not listen to the rule of a blood alcohol level of 0.04.

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