/*! 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 40 Solve each exponential equation.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve each exponential equation. Express the solution set in terms of natural logarithms or common logarithms. Then use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$3^{\frac{x}{7}}=0.2$$

Short Answer

Expert verified
The short answer for x, correct to two decimal places, would be obtained from the calculation in Step 4. The result should be written here after performing the calculations.

Step by step solution

01

Transform the exponential equation into a logarithmic equation

By transforming, the initial equation \[3^{\frac{x}{7}}=0.2\] can be rewritten as \[\log_{3}0.2=\frac{x}{7}\] using logarithmic properties.
02

Isolate the variable

Isolate x by multiplying both sides by 7, therefore \[x=7\log_{3}0.2\].
03

Calculate decimal approximation

Now, use a calculator to obtain a decimal approximation. It should be mentioned that the natural logarithm might be more available on standard calculators. Conversion can be made using the formula \[ \log_{a}b=\frac{\ln{b}}{\ln{a}} \]}, and this gives \[ x=7\frac{\ln{0.2}}{\ln{3}} \]
04

Obtain the final decimal approximation

Using the calculator gives the final decimal approximation, correct to two decimal places.

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 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The logistic growth function $$ P(x)=\frac{90}{1+271 e^{-0.122 x}} $$ models the percentage, \(P(x),\) of Americans who are \(x\) years old with some coronary heart disease. Use the function to solve Exercises \(43-46\) What percentage of 20 -year-olds have some coronary heart disease?

The logistic growth function $$ P(x)=\frac{90}{1+271 e^{-0.122 x}} $$ models the percentage, \(P(x),\) of Americans who are \(x\) years old with some coronary heart disease. Use the function to solve Exercises \(43-46\) What percentage of 80 -year-olds have some coronary heart disease?

Hurricanes are one of nature's most destructive forces. These low-pressure areas often have diameters of over 500 miles. The function \(f(x)=0.48 \ln (x+1)+27\) models the barometric air pressure, \(f(x),\) in inches of mercury, at a distance of \(x\) miles from the eye of a hurricane. The function \(W(t)=2600\left(1-0.51 e^{-0.075 t}\right)^{3}\) models the weight, \(W(t),\) in kilograms, of a female African elephant at age \(t\) years. (1 kilogram \(=2.2\) pounds) Use a graphing utility to graph the function. Then \([\mathrm{TRACE}]\) along the curve to estimate the age of an adult female elephant weighing 1800 kilograms.

We see from the calculator screen at the bottom of the previous page that a logistic growth model for world population, \(f(x),\) in billions, \(x\) years affer 1949 is $$ f(x)=\frac{12.57}{1+4.11 e^{-0.026 x}} $$ Use this function to solve Exercises \(38-42\) How well does the function model the data showing a world population of 6.1 billion for \(2000 ?\)

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I can solve \(4^{x}=15\) by writing the equation in logarithmic form.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.