/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Rewrite the given expression wit... [FREE SOLUTION] | 91Ó°ÊÓ

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Rewrite the given expression without using any exponentials or logarithms. $$ \left(e^{3}\right)^{\ln (4)}-\left(3^{\ln (59)}\right)^{\log _{3}(e)} $$

Short Answer

Expert verified
The expression simplifies to 5.

Step by step solution

01

Apply Exponential and Logarithmic Properties

The expression \(ig(e^{3}\big)^{\ln(4)}\) can be rewritten using the property \((a^b)^c = a^{bc}\). Hence, \[\big(e^{3}\big)^{\ln(4)} = e^{3 \cdot \ln(4)}.\] Similarly, \(\big(3^{\ln(59)}\big)^{\log_3(e)}\) can be rewritten as \[3^{\ln(59) \cdot \log_3(e)}.\]
02

Use Change of Base Formula

The logarithmic term \(\log_{3}(e)\) can be rewritten using the change of base formula: \(\log_3(e) = \frac{\ln(e)}{\ln(3)}.\) Given that \(\ln(e) = 1\), it simplifies to \(\frac{1}{\ln(3)}\). Thus, \[3^{\ln(59) \cdot \log_3(e)} = 3^{\ln(59) \cdot \frac{1}{\ln(3)}} = 3^{\frac{\ln(59)}{\ln(3)}}.\]
03

Simplify Exponents

Using the property \(a^{\frac{\ln(b)}{\ln(a)}} = b\), we apply it to the expression \(3^{\frac{\ln(59)}{\ln(3)}}\). Consequently, \(3^{\frac{\ln(59)}{\ln(3)}} = 59\). In similar manner, for \(e^{3 \ln (4)}\), this simplifies to \((e^{\ln 4})^3 = 4^3 = 64\).
04

Subtract the Simplified Values

Now, subtract the simplified value from Step 3: \(64 - 59 = 5\). The rewritten expression is \(64 - 59 = 5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Change of Base Formula
The change of base formula is a powerful tool that helps convert logarithms from one base to another. Many students find this concept a bit tricky at first, but once you get the hang of it, it simplifies a lot of complex expressions. The formula is given by:
\[\log_b(a) = \frac{\log_k(a)}{\log_k(b)}\]where \( b \) and \( k \) are the old and new bases, respectively. In most cases, \( k \) is often 10 (common logarithm) or \( e \) (natural logarithm). The base \( e \) is particularly handy in calculus and higher-level mathematics.
  • In our solution, the base was changed using \( \log_3(e) = \frac{\ln(e)}{\ln(3)} \).
  • The natural logarithm of \( e \), \( \ln(e) \), is 1.
Applying this formula helps us to reframe complex problems, such as converting between different logarithmic bases to simplify calculations or make further mathematical manipulations possible. **Understanding the change of base formula is essential for ease in handling logarithmic expressions.**
Exponential Properties
Exponents are shorthand for expression simplification and play a crucial role in algebra and calculus. To grasp exponential properties well, it's essential to understand their basics. The key properties include:
  • Product of Powers: \( a^m \times a^n = a^{m+n} \)
  • Power of a Power: \( (a^m)^n = a^{m\cdot n} \)
  • Power of a Product: \( (ab)^n = a^n \cdot b^n \)
In the given exercise, the **Power of a Power property** is applied. For example, \( \left(a^{b}\right)^{c} \) can be rewritten as \( a^{b \cdot c} \). This can drastically simplify expressions and solutions.
In the exercise, we simplify \( \left(e^{3}\right)^{\ln(4)} \) to \( e^{3 \cdot \ln(4)} \). This made the original expression much easier to handle, revealing a clearer path to the solution.
Logarithmic Properties
Logarithmic properties help us to manipulate logarithmic expressions creatively. These properties relate deeply to exponential functions, giving us alternative ways to express and break down complex problems.
Key logarithmic properties include:
  • Product Property: \( \log_b(MN) = \log_b(M) + \log_b(N) \)
  • Quotient Property: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
  • Power Property: \( \log_b(M^p) = p\cdot \log_b(M) \)
In this exercise, **logarithmic properties** enable the conversion and simplification of expressions. For example, the property \( a^{\frac{\ln(b)}{\ln(a)}} = b \) is used to transform expressions into a more simplified form directly.
Understanding these properties provides critical insight into solving logarithmic and exponential problems. They allow us to move fluidly between these forms, presenting a more systematic and efficient path to answers.

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Most popular questions from this chapter

A particle moves on an axis. Its position \(p(t)\) at time \(t\) is given. For a positive \(h,\) the average velocity over the time interval \([2,2+h]\) is \(\bar{v}(h)=\frac{p(2+h)-p(2)}{h}\) a. Numerically determine \(v_{0}=\lim _{h \rightarrow 0+} \bar{v}(h)\). b. How small does \(h\) need to be for \(\bar{v}(h)\) to be between \(v_{0}\) and \(v_{0}+0.1 ?\) c. How small does \(h\) need to be for \(\bar{v}(h)\) to be between \(v_{0}\) and \(v_{0}+0.01 ?\) $$ p(t)=2 t-\frac{24}{\pi} \cos \left(\frac{\pi t}{12}\right) $$

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In each of Exercises \(93-96,\) plot the given functions \(g, f,\) and \(h\) in a common viewing rectangle that illustrates \(f\) being pinched at the point \(c\). Determine \(\lim _{x \rightarrow c} f(x)\). $$ \begin{array}{l} g(x)=2, f(x)=2+(x-1)^{2} \cos ^{2}(x /(x-1)) \\\ h(x)=2+|x-1| / 4 \quad c=1 \end{array} $$

In each of Exercises \(93-96,\) plot the given functions \(g, f,\) and \(h\) in a common viewing rectangle that illustrates \(f\) being pinched at the point \(c\). Determine \(\lim _{x \rightarrow c} f(x)\). $$ \begin{array}{l} g(x)=-2|x|, f(x)=\cos (x-1 / x)-\cos (x+1 / x) \\\ h(x)=2|x| \quad c=0 \end{array} $$

In each of Exercises \(93-96,\) plot the given functions \(g, f,\) and \(h\) in a common viewing rectangle that illustrates \(f\) being pinched at the point \(c\). Determine \(\lim _{x \rightarrow c} f(x)\). $$ \begin{array}{l} g(x)=2-4|x|, f(x)=\left(2+x^{2}\right) /\left(1+x^{2}\right)+\sin (x+1 / x)+ \\ \sin (x-1 / x), h(x)=2(1+|x|) \quad c=0 \end{array} $$

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