/*! 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 24 Solve the equation. Write the so... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(2^{z}=70\)

Short Answer

Expert verified
The exact solution is \(\frac{\ln(70)}{\ln(2)}\). The approximate solution is 6.1297.

Step by step solution

01

Write the equation

Start with the given equation: \[2^{z}=70\]
02

Apply the natural logarithm

Take the natural logarithm (\ln) of both sides of the equation:\[\ln(2^{z}) = \ln(70)\]
03

Use the power rule of logarithms

Apply the power rule of logarithms, which states \(\ln(a^{b}) = b \ln(a)\) to move the exponent outside the logarithm: \[ z \ln(2) = \ln(70)\]
04

Solve for z

Isolate z by dividing both sides of the equation by \(\ln(2)\): \[ z = \frac{\ln(70)}{\ln(2)}\]
05

Simplify the logarithms

Use a calculator to find the values of the natural logarithms: \(\ln(70) \approx 4.2485\) and \(\ln(2) = 0.6931\). Then divide these values: \[z \approx \frac{4.2485}{0.6931} \approx 6.1297\]

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Key Concepts

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

Natural Logarithms
A natural logarithm, denoted as \(\text{ln}\), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. In simpler terms, the natural logarithm of a number is the power to which \(\text{e}\) must be raised to obtain that number. For example, \(\text{ln}(70)\) means the power to which we need to raise \(\text{e}\) to get 70.
Natural logarithms are often used because they simplify many mathematical operations, especially when dealing with exponential growth and decay processes. Compared to common logarithms, which use base 10, natural logarithms are more prevalent in advanced mathematics and physics.
Applying natural logarithms to both sides of an equation, such as \(2^{z}=70\), helps in transforming an exponential equation into a linear form, making it easier to solve.
Power Rule of Logarithms
The power rule of logarithms is a critical tool for solving logarithmic equations. The rule states that for any positive real numbers \(a\) and \(b\), and a real number \(c\), \(\text{ln}(a^b)=b \text{ln}(a)\). This rule allows us to move the exponent in a logarithmic expression to become a coefficient in front of the logarithm.
For example, in the equation \(2^z = 70\), applying the natural logarithm to both sides gives us \(\text{ln}(2^z) = \text{ln}(70)\). Using the power rule, this transforms into \(z \text{ln}(2) = \text{ln}(70)\), substantially simplifying the problem by removing the exponent.
Exact Solutions
An exact solution to an equation is one that is expressed in a precise mathematical form without any approximations. For exponential equations like \(2^z=70\), the exact solution involves using logarithms. After applying the natural logarithms and using the power rule, we isolate \(\text{z}\) to find \(\text{z} = \frac{\text{ln}(70)}{\text{ln}(2)}\).
This fraction represents our exact solution in terms of natural logarithms. It is exact because it provides the relationship between the variables without rounding any numbers.
Approximate Solutions
Sometimes, it’s useful to find approximate solutions to an equation for practical purposes or ease of use. After finding the exact solution \(z = \frac{\text{ln}(70)}{\text{ln}(2)}\), we can use a calculator to find approximate decimal values for the logarithms.
Using a calculator, \(\text{ln}(70) \approx 4.2485\) and \(\text{ln}(2) \approx 0.6931\). Dividing these results gives us \(z \approx \frac{4.2485}{0.6931} \approx 6.1297\).
This value is rounded to four decimal places, providing a practical approximation of the solution to the problem.

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a. The populations of two countries are given for January 1,2000 , and for January 1,2010 . Write a function of the form \(P(t)=P_{0} e^{k t}\) to model each population \(P(t)\) (in millions) \(t\) years after January 1, 2000.$$ \begin{array}{|l|c|c|c|} \hline & \begin{array}{c} \text { Population } \\ \text { in 2000 } \\ \text { (millions) } \end{array} & \begin{array}{c} \text { Population } \\ \text { in 2010 } \\ \text { (millions) } \end{array} & \boldsymbol{P}(t)=\boldsymbol{P}_{0} e^{k t} \\ \hline \text { Switzerland } & 7.3 & 7.8 & \\ \hline \text { Israel } & 6.7 & 7.7 & \\ \hline \end{array}$$ b. Use the models from part (a) to predict the population on January \(1,2020,\) for each country. Round to the nearest hundred thousand. c. Israel had fewer people than Switzerland in the year 2000 , yet from the result of part (b), Israel will have more people in the year \(2020 ?\) Why? d. Use the models from part (a) to predict the year during which each population will reach 10 million if this trend continues.

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