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Chapter 3: Problem 60

Rewrite the equation in terms of base \(e\). Express the answer in terms of a natural logarithm and then round to three decimal places. $$y=4.5(0.6)^{x}$$

Short Answer

Expert verified
The given equation in terms of base \(e\) in the natural logarithm form and rounded to three decimal places (where applicable) is \(x = ln(y/4.5) / ln(0.6)\)

Step by step solution

01

Express the given exponent in terms of base e

The equation given is \(y = 4.5(0.6)^x\). We can rewrite the 0.6 as \(e^{ln(0.6)}\) because \(e^{ln(a)}=a\). So this becomes \(y = 4.5e^{x*ln(0.6)}\).
02

Rewrite into logarithmic form

We now rewrite this into logarithmic form. This can be done using the power rule for logarithms which states that \(e^{ln(a)} = a\). Therefore, we obtain \(x = ln(y/4.5) / ln(0.6)\).
03

Round to three decimal places

Finally, we should give our answer as a natural logarithm and round the answer to three decimal places to retain precision. However, to keep the solution general, we leave \(y\) as it is without any specific value. When a specific value of \(y\) is given, it will be substituted into \(x = ln(y/4.5) / ln(0.6)\) and the result will be rounded to three decimal places.

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Most popular questions from this chapter

Explain the differences between solving \(\log _{3}(x-1)=4\) and \(\log _{3}(x-1)=\log _{3} 4\)

In parts (a)-(c), graph \(f\) and \(g\) in the same viewing rectangle. a. \(f(x)=\ln (3 x), g(x)=\ln 3+\ln x\) b. \(f(x)=\log \left(5 x^{2}\right), g(x)=\log 5+\log x^{2}\) c. \(f(x)=\ln \left(2 x^{3}\right), g(x)=\ln 2+\ln x^{3}\) d. Describe what you observe in parts (a)-(c). Generalize this observation by writing an equivalent expression for \(\log _{b}(M N),\) where \(M>0\) and \(N>0\) e. Complete this statement: The logarithm of a product is equal to____.

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I estimate that \(\log _{8} 16\) lies between 1 and 2 because \(8^{1}=8\) and \(8^{2}=64\).

Find the inverse of \(f(x)=x^{2}+4, x \geq 0\) (Section \(1.8, \text { Example } 7)\).

This will help you prepare for the material covered in the first section of the next chapter. $$\text { Simplify: } \frac{17 \pi}{6}-2 \pi$$
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