Chapter 4: Problem 19
Solve logarithmic equation. \(x=3^{\log _{3} 8}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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}
Learning Materials
Features
Discover
Chapter 4: Problem 19
Solve logarithmic equation. \(x=3^{\log _{3} 8}\)
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
For individual or collaborative investigation (Exercises \(117-122\) ) Assume \(f(x)=a^{x}\), where \(a>1 .\) Work these exercises in order. If \(\left.a=10, \text { what is the equation for } y=f^{-1}(x) ? \text { (You need not solve for } y .\right)\)
Use a graphing calculator to find the solution set of each equation. Approximate the solution \((s)\) to the nearest tenth. $$x=2^{x}$$
Medication Effectiveness Drug effectiveness decreases over time. If, each hour, a drug is only \(90 \%\) as effective as the previous hour, at some point the patient will not be receiving enough medication and must receive another dose. If the initial dose was \(200 \mathrm{mg}\) and the drug was administered \(3 \mathrm{hr}\) ago, the expression \(200(0.90)^{3},\) which equals \(145.8,\) represents the amount of effective medication still in the system. (The exponent is equal to the number of hours since the drug was administered.)
In calculus, it is shown that $$ e^{x}=1+x+\frac{x^{2}}{2 \cdot 1}+\frac{x^{3}}{3 \cdot 2 \cdot 1}+\frac{x^{4}}{4 \cdot 3 \cdot 2 \cdot 1}+\frac{x^{5}}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}+\cdots $$ By using more terms, one can obtain a more accurate approximation for \(e^{x}\). Use the terms shown, and replace \(x\) with 1 to approximate \(e^{1}=e\) to three decimal places. Check your result with a calculator.
Use the change-of-base theorem to find an approximation to four decimal places for each logarithm. $$\log _{0.32} 5$$
What do you think about this solution?
We value your feedback to improve our textbook solutions.