/*! 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 66 Solve each equation for \(x\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve each equation for \(x\) $$ \text { (a) }\ln (\ln x)=1 \quad \text { (b) } e^{a x}=C e^{b x}, \text { where } a \neq b $$

Short Answer

Expert verified
(a) \( x = e^e \); (b) \( x = \frac{\ln C}{a-b} \).

Step by step solution

01

Solving the equation for part (a)

The given equation is \( \ln (\ln x) = 1 \). To solve for \( x \), we first exponentiate both sides to remove the outer logarithm, leading to \( \ln x = e^1 = e \).
02

Finding \( x \) value for part (a)

Now, remove the logarithm by exponentiating again, obtaining \( x = e^e \). Thus, \( x \) is \( e^e \) for part (a).
03

Isolating exponentials for part (b)

The given equation is \( e^{ax} = C e^{bx} \). Dividing both sides by \( e^{bx} \) gives \( e^{ax-bx} = C \), which simplifies to \( e^{(a-b)x} = C \).
04

Solving for \( x \) in part (b)

Take the natural logarithm of both sides to get \( (a-b)x = \ln C \). Solving for \( x \) gives the solution: \( x = \frac{\ln C}{a-b} \) provided \( a eq b \).

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Key Concepts

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

Exponential Equations
An exponential equation is one in which variables appear as exponents. These types of equations often arise in scientific, financial, and growth-related models. The essence of such equations lies in balancing the powers relative to a base, typically denoted as 'e' in natural sciences. In the given example, the equation is in the form of \[ e^{ax} = C e^{bx} \]where the concept of balancing exponentials is visibly at play. To begin solving, divide both sides by \( e^{bx} \), simplifying the equation to \[ e^{(a-b)x} = C \]Now, the key is to isolate \(x\) by employing natural logarithms.By equating the powers, you effectively turn the exponent's complexity into a manageable algebraic expression, as in\[ (a-b)x = \ln C \] This method helps in streamlining the solution for \( x \), providing straightforward results once you overcome the initial exponent phase. The common approach involves using properties of exponentials and logarithms, making such problems easier to visualize and solve.
Natural Logarithm
The natural logarithm, denoted by \( \ln \), is the logarithm to the base \( e \), an irrational constant approximately equal to 2.71828. It is often used in various branches of mathematics due to its natural emergence in calculus and continuous growth models.In tackling the equation \( \ln (\ln x) = 1 \), knowing the properties of the natural logarithm, specifically its inverse relationship with exponentials, becomes crucial. By exponentiating both sides, \[ \ln x = e^1 = e \] we unfold the logarithmic complexity into a simpler form. In essence, the natural logarithm helps in managing multiplicative processes and growth scenarios conveniently, allowing for linear clarity when solving for variables inside the logarithm.
Exponential Function
The exponential function, often expressed as \( e^x \), is a cornerstone in mathematical equations involving growth or decay. This function is characterized by a constant base of \( e \), allowing it to model continuous growth efficiently. You encounter the exponential function regularly in many scientific domains, from population studies to radioactive decay.In the context of solving the equation in part (a), the repeated application of exponential functions serves as the tool for unwrapping nested logarithms. Once we've identified that \( \ln x = e \) from the natural logarithm rules, invoking the exponential function gives \[ x = e^e \] This demonstrates the efficiency of exponential functions in translating complex logarithmic forms into real-world numbers.The universality of the exponential function makes it indispensable when dealing with exponential growth or decay, especially when computations necessitate precision and simplicity.

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Most popular questions from this chapter

Express the given quantity as a single logarithm. $$ \ln 5+5 \ln 3 $$

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$$f(x)=\left\\{\begin{array}{ll}{|x|^{x}} & {\text { if } x \neq 0} \\ {1} & {\text { if } x=0}\end{array}\right.$$ (a) Show that \(f\) is continuous at 0 . (b) Investigate graphically whether \(f\) is differentiable at 0 by zooming in several times toward the point \((0,1)\) on the graph of \(f .\) (c) Show that \(f\) is not differentiable at \(0 .\) How can you reconcile this fact with the appearance of the graphs in part (b)?

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