Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 12
Solve for \(x\) by changing to exponential form. Round your answer to three decimal places. a. \(\ln 3 x=1\) b. \(3 \ln x=5\) c. \(\ln 3+\ln x=1.5\)
Short Answer
Expert verified
a. \(x \approx 0.906\), b. \(x \approx 5.294\), c. \(x \approx 1.494\).
Step by step solution
01
Solve for part (a) \(\ln 3x = 1\)
To solve for x, we first need to change the logarithmic form to exponential form. The logarithmic form \(\ln 3x = 1\) can be written as \(\text{e}^1 = 3x\).
02
Isolate x in part (a)
Divide both sides by 3 to isolate x: \(\frac{\text{e}^1}{3} = x\). This simplifies to \(x = \frac{\text{e}}{3}\).
03
Calculate the value of x for part (a)
Compute the value using a calculator: \(x = \frac{2.718}{3} \approx 0.906\). Round the answer to three decimal places: \(x \approx 0.906\).
04
Solve for part (b) \ (3 \ln x = 5)\
Divide both sides by 3 to isolate \(\ln x\): \(\ln x = \frac{5}{3}\).
05
Change to exponential form for part (b)
Rewrite the equation in exponential form: \(\text{e}^{\frac{5}{3}} = x\).
06
Calculate the value of x for part (b)
Compute the value using a calculator: \(x \approx \text{e}^{1.667} \approx 5.294\). Round the answer to three decimal places: \(x \approx 5.294\).
07
Solve for part (c) \(\ln 3 + \ln x = 1.5\)
Combine the logarithms using the product property: \(\text{ln}(3x) = 1.5\).
08
Change to exponential form for part (c)
Rewrite the equation in exponential form: \(3x = \text{e}^{1.5}\).
09
Isolate x in part (c)
Divide both sides by 3 to isolate x: \(x = \frac{\text{e}^{1.5}}{3}\).
10
Calculate the value of x for part (c)
Use a calculator to find \(x = \frac{4.481}{3} \approx 1.494\). Round the answer to three decimal places: \(x \approx 1.494\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
logarithmic form
Logarithmic equations help us solve for variables hidden within a logarithm. The general form of a logarithmic equation is given by \(\text{log}_b(y) = x\), where \(b\) is the base of the logarithm, \(y\) is the argument, and \(x\) is the exponent or the result. Understanding how to manipulate these forms is essential in solving logarithmic equations. For example, in part (a) \(\text{ln }3x = 1\), \(\ln\) represents the natural logarithm, which has a base \(e\). We need to transition from this form to its exponential counterpart to solve for \(x\).
exponential form
Converting from logarithmic to exponential form makes solving easier. If you have \(\text{log}_b(y) = x\), it can be rewritten in exponential form as \(b^x = y\). Let's see an example: For the equation \(\ln 3x = 1\), the natural logarithm with base \(e\) converts directly to \(e^1 = 3x\). This is a critical step, as it allows us to work with exponentiation rather than logarithms, making it easier to isolate the variable.
isolating variables
Isolating variables is a fundamental algebraic skill, necessary for solving equations irrespective of their form鈥攍inear, logarithmic, or exponential. In the case of \(e^1 = 3x\) from part (a), we isolate \(x\) by dividing both sides by 3, giving \(\frac{e^1}{3} = x\). Similarly, for part (b) \(3 \ln x = 5\), we divide both sides by 3 to solve for \(\ln x\): \(\ln x = \frac{5}{3}\). This step simplifies our work and directly leads to the solution.
rational exponents
Rational exponents reflect fractional powers and can frequently appear when dealing with exponential equations. For instance, when converting back to the exponential form, like \(\ln x = \frac{5}{3}\) becoming \(e^\frac{5}{3} = x\), we see a rational exponent in action. Using a calculator to compute \(e^\frac{5}{3}\), we can readily find \(\text{e}^{1.667} \approx 5.294\). Rational exponents are crucial in rendering complex exponential equations accessible and more straightforward to solve.
