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Chapter 5: Problem 83

$\mathrm{S}=30\left(1-\mathrm{e}^{(\ln 5 / 6) 5}\right)=30\left(1-\left(\frac{5}{6}\right)^{5}\right)$

Short Answer

Expert verified
Since this is a multiplication and a natural logarithm, we can start by finding the logarithm of 5: $$ \ln 5 \approx 1.6094 $$ Now, divide this result by 6: $$ \frac{1.6094}{6} \approx 0.2682 $$ Next, multiply the result by 5: $$ 0.2682 \times 5 \approx 1.341 $$ #tag_title# Step 3: Evaluate the exponential term#tag_content# Now that we have the exponent, we can evaluate the e term: $$ e^{1.341} \approx 3.823 $$ This means that our expression is now essentially: $$ S = 30(1 - 3.823) $$ #tag_title# Step 4: Complete the subtraction and multiplication#tag_content# Finally, we can complete the subtraction inside the parentheses and then multiply with the factor 30: $$ 1 - 3.823 = -2.823 $$ Multiply this result by 30: $$ S = 30 \times (-2.823) \approx -84.69 $$ #tag_answer# The result of the expression is approximately -84.69.

Step by step solution

01

Identify the terms and operations

In this expression, we have several operations: exponentiation, subtraction, multiplication, and a natural logarithm. We'll deal with them one by one, following the order of operations. The given expression is: $$ S = 30(1 - e^{(\ln 5 / 6)5}) = 30(1 - (\frac{5}{6})^5) $$
02

Evaluate the exponent with the natural logarithm

First, we'll focus on the exponent inside of the e term: $$ (\ln 5 / 6) 5 $$

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Most popular questions from this chapter

\(f(x)=\frac{x}{1-x^{2}}\) $f^{\prime}(x)=\frac{\left(1-x^{2}\right)-x(-2 x)}{\left(1-x^{2}\right)^{2}}=\frac{1+x^{2}}{\left(1-x^{2}\right)^{2}}=1$ \(1+x^{2}=1+x^{4}-2 x^{2}\) \(\Rightarrow \quad x^{2}\left(x^{2}-3\right)=0\) \(\Rightarrow x=0, \pm \sqrt{3}\) $P t \rightarrow(0,0),\left(\sqrt{3}, \frac{\sqrt{3}}{2}\right)\left(-\sqrt{3}, \frac{\sqrt{3}}{2}\right)$

\(\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2\) $\frac{n}{a}\left(\frac{x}{a}\right)^{n-1}+\frac{n}{b}\left(\frac{y}{b}\right)^{n-t} \frac{d y}{d x}=0$ \(\frac{d y}{d x}=-\frac{b^{n} x^{n-1}}{a^{n} y^{n-1}}=-\frac{b}{a}\)

\(f_{1}^{\prime}(x)=2 x-1 \quad \& \quad f_{2}^{\prime}(x)=3 x^{2}-2 x-2\) \(\Rightarrow 2 x_{1}-1=3 x_{2}^{2}-2 x_{2}-2\) \(\Rightarrow 3 x_{2}^{2}-2 x_{2}-2 x_{1}-1=0\) For \(\mathrm{x}_{2}\) be real, \(\mathrm{D} \geq 0\) \(4-4\left(2 x_{1}+1\right)(3)\) \(4-24 x_{1}-12\) \(\Rightarrow-\left(24 x_{1}+8\right)\) There can be infinite such values of \(\mathrm{x}_{\mathrm{l}}\)

\(y=-t+e^{a t}=0\) \(x=t+e^{a}\) \(\frac{d y}{d t}=-1+a e^{u r}\) \(\frac{d x}{d t}=1+a e^{u}\) \(\frac{d y}{d x}=\frac{a e^{a t}-1}{a e^{a t}+1}=0\) \(\Rightarrow \mathrm{e}^{a t}=\frac{1}{a}\) \(\Rightarrow a t=-\ln a\) \(\mathrm{t}=-\frac{\ln \mathrm{a}}{\mathrm{a}}\) $\mathrm{x}=-\frac{\ln \mathrm{a}}{\mathrm{a}}+\frac{1}{\mathrm{a}}=\frac{\ln \mathrm{e} / \mathrm{a}}{\mathrm{a}}=\frac{2}{\mathrm{a}}$

\(x y^{2}=1\) \(y^{2}+2 x y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-y}{2 x}\) \(-\frac{d x}{d y}=\frac{2 x}{y}=\frac{2}{y^{3}}\) \(y-y_{1}=\frac{2}{y_{1}^{3}}\left(x-x_{1}\right)\) \(+y_{1}^{4}=2 x_{1}\) \(y_{1}^{6}=2\) \(y_{1}=\pm 2^{1 / 6}\) \(x_{1}=\pm 2^{-\sqrt{3}}\)
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