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Exer. 11-28: Find the exact solution, using common logarithms, and a two- decimal-place approximation of each solution, when appropriate. $$4^{2 x+3}=5^{x-2}$$

Short Answer

Expert verified
The exact solution: \(x = \frac{-2 \log(5) - 3 \log(4)}{2 \log(4) - \log(5)}\); approx. \(x \approx -6.34\).

Step by step solution

01

Apply the Logarithm to Both Sides

To solve the equation \(4^{2x+3} = 5^{x-2}\), we first apply the common logarithm (log base 10) to both sides of the equation to bring the exponents down. This gives us: \[\log(4^{2x+3}) = \log(5^{x-2})\] Using the property of logarithms that \(\log(a^b) = b \cdot \log(a)\), we rewrite this as: \[(2x+3) \cdot \log(4) = (x-2) \cdot \log(5)\]
02

Expand the Logarithmic Equation

Expand both sides of the equation using the distributive property: \[2x \cdot \log(4) + 3 \cdot \log(4) = x \cdot \log(5) - 2 \cdot \log(5)\] This expansion will help us isolate the variable \(x\).
03

Rearrange the Equation to Isolate x

Rearrange the equation to bring all terms involving \(x\) to one side:\[2x \cdot \log(4) - x \cdot \log(5) = -2 \cdot \log(5) - 3 \cdot \log(4)\]This step sets the stage for factoring out \(x\).
04

Factor and Solve for x

Factor out \(x\) from the left side of the equation:\[x (2 \cdot \log(4) - \log(5)) = -2 \cdot \log(5) - 3 \cdot \log(4)\]Now, solve for \(x\) by dividing both sides by \(2 \cdot \log(4) - \log(5)\): \[x = \frac{-2 \cdot \log(5) - 3 \cdot \log(4)}{2 \cdot \log(4) - \log(5)}\]
05

Calculate the Result

Use a calculator to evaluate the logs and find the approximate value of \(x\). First, calculate:\[\log(4) \approx 0.60206, \quad \log(5) \approx 0.69897\]Substitute these values into the expression for \(x\):\[x \approx \frac{-2 \cdot 0.69897 - 3 \cdot 0.60206}{2 \cdot 0.60206 - 0.69897} \approx \frac{-1.39794 - 1.80618}{1.20412 - 0.69897}\]\[x \approx \frac{-3.20412}{0.50515} \approx -6.34\]Thus, the approximate solution is \(x \approx -6.34\).

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Key Concepts

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

Common Logarithms
The term 'common logarithms' refers to logarithms that use a base of 10. They are often denoted simply as 'log'. This is the default logarithm for many calculators, making it practical for solving equations and performing calculations in science and engineering.

The primary role of common logarithms is to simplify equations involving exponential expressions. For example, taking the logarithm of both sides of an equation can help transform an exponent into a coefficient, making it easier to solve. This is seen in the solution to the exercise where we transform the equation \(4^{2x+3} = 5^{x-2}\) by applying the common logarithm:
  • The left side becomes \(\log(4^{2x+3})\), which simplifies to \((2x+3) \cdot \log(4)\) using properties of logarithms.
  • The right side becomes \(\log(5^{x-2})\), simplifying to \((x-2) \cdot \log(5)\).
This step transforms the equation into a form where algebraic methods can be used to solve for \(x\). This demonstrates the power of common logarithms in dealing with exponential equations.
Exponential Functions
Exponential functions are a crucial concept in mathematics and occur frequently in various applied domains, such as physics and finance. These functions have the form \(a^{f(x)}\), where \(a\) is a positive constant called the base and \(f(x)\) is the exponent. The exponential function is characterized by rapid growth or decay.

In our exercise, we encounter an exponential equation: \(4^{2x+3} = 5^{x-2}\). Here, both sides exhibit different bases (4 and 5), which influences how the equation behaves and needs to be managed. Addressing equations with differing exponential bases involves rewriting them using logarithms.
  • The rapid growth or decay of exponential functions means small changes in \(x\) can lead to significant changes in the output, making equations like this non-trivial to solve without the assistance of logarithms.
  • Understanding the behavior of these functions helps predict how variables will interact in the equation and how solutions will be plotted on a graph.
Therefore, one must be equipped with tools and techniques like logarithms to effectively solve and analyze exponential functions.
Logarithmic Properties
Logarithmic properties are indispensable tools when working with equations involving logarithms. These properties allow for the manipulation and simplification of logarithmic expressions and are essential for solving complex equations like the one in the exercise.

Key logarithmic properties include:
  • \(\log(a^b) = b \cdot \log(a)\): This property simplifies exponentiation into multiplication, which makes it easier to solve exponential equations numerically.
  • \(\log(ab) = \log(a) + \log(b)\): Useful for expanding or combining logarithmic expressions.
  • \(\log(a/b) = \log(a) - \log(b)\): Helps simplify division inside a logarithm to subtraction, streamlining the equation further.
In the exercise solution, the property \(\log(a^b) = b \cdot \log(a)\) converts the initial exponential form into a linear form, making the variable \(x\) more accessible for algebraic operations.

Mastery of these properties allows for greater flexibility and efficiency in solving equations involving exponentials and logarithms, and they are vital for understanding the mechanics of these mathematical tools.

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Most popular questions from this chapter

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