/*! 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 20 Solve the equations. $$ 10^{... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 20

Solve the equations. $$ 10^{t}-5.4=7.2-10^{t} $$

Short Answer

Expert verified
Question: Solve the equation \(10^t - 5.4 = 7.2 - 10^t\) for t. Answer: The solution to the given equation is approximately \(t = 0.799\).

Step by step solution

01

Rewrite the given equation

Rewrite the given equation as: $$ 10^t - 5.4 = 7.2 - 10^t $$
02

Move the term with the variable to one side of the equation

To do this, we add \(10^t\) to both sides of the equation: $$ 10^t - 5.4 + 10^t = 7.2 - 10^t + 10^t $$ This simplifies to: $$ 2 \cdot 10^t - 5.4 = 7.2 $$
03

Isolate the term with the variable

Now we isolate the term with the variable by adding 5.4 to both sides of the equation: $$ 2 \cdot 10^t - 5.4 + 5.4 = 7.2 + 5.4 $$ This simplifies to: $$ 2 \cdot 10^t = 12.6 $$
04

Divide by the coefficient of the term with the variable

In this step, we will divide both sides by 2 to further isolate the term with the variable: $$ \frac{2 \cdot 10^t}{2} = \frac{12.6}{2} $$ This simplifies to: $$ 10^t = 6.3 $$
05

Take the logarithm on both sides of the equation

To solve for t, we can take the logarithm of both sides. Let's take the base 10 logarithm (logarithm with base 10 is usually represented as "log"): $$ \log{10^t} = \log{6.3} $$
06

Use logarithm properties to simplify the equation

Now we'll apply the logarithm property that states \(\log{a^b} = b\log{a}\): $$ t\log{10} = \log{6.3} $$ Since \(\log{10} = 1\), we have: $$ t = \log{6.3} $$
07

Calculate the value of t

Now, we can use a calculator to find the value of t: $$ t \approx 0.799 $$ Thus, the solution to the given equation is approximately \(t = 0.799\).

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Key Concepts

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

Logarithms
When solving exponential equations, a crucial tool we often use is logarithms.
Logarithms can help transform exponential forms into linear forms, making it easier to solve for unknowns.
In the equation \(10^t = 6.3\), using logarithms is key to finding the value of \(t\).

Understanding Logarithms

Logarithms answer the question: "To what exponent must we raise the base to get a certain value?"
For example, in \(\log_{10} 100\), we are asking: "What power of 10 equals 100?" The answer is 2, because \(10^2 = 100\).
This property of logarithms allows us to peel back the exponential in equations.

Applying Logarithms to Solve Equations

In the step \(\log(10^t) = \log(6.3)\), we use the property \(\log(a^b) = b \log(a)\) to bring down the exponent \(t\).
This simplifies our equation to \(t \cdot \log(10) = \log(6.3)\).
Since \(\log(10) = 1\), the equation becomes \(t = \log(6.3)\), which can be easily calculated.
Isolating Variables
Isolating variables is a critical step in solving algebraic equations.
The goal is to get the variable by itself on one side of the equation.
This makes it easier to find its value. Let's look at how this works step by step.

Moving Terms

Our example starts with the equation \(10^t - 5.4 = 7.2 - 10^t\).
The variable \(10^t\) appears on both sides. To isolate it, we move all terms containing \(t\) to one side.
This is done by adding \(10^t\) to both sides, resulting in \(2 \cdot 10^t - 5.4 = 7.2\).

Eliminating Constants

Next, we add or subtract constants to further isolate the term with the variable.
In this case, adding 5.4 to both sides gives \(2 \cdot 10^t = 12.6\).
Now, \(10^t\) is almost by itself, ready for the final step.
Solving Algebraic Equations
Solving algebraic equations involves systematically working through steps to find the value of unknowns.
It can include operations like addition, subtraction, division, and the use of logarithms.

Breaking Down the Equation

Start by simplifying the equation. Move similar terms together. In our example, this meant gathering all \(t\)-terms.
Always work to get the variable alone by performing inverse operations.
For example, if \(10^t\) is multiplied by 2, divide both sides by 2 to counteract this.

Final Steps

Once isolated, we used logarithms on both sides in \(10^t = 6.3\).
This converted the equation to \(t = \log(6.3)\), which is easily solvable using a calculator.
This structured process allows anyone to solve similar exponential equations effectively.

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Most popular questions from this chapter

In \(1936,\) researchers at the Wright-Patterson Air Force Base found that each new aircraft took less time to produce than the one before, owing to what is now known as the learning curve effect. Specifically, researchers found that \(^{16}\) $$ \begin{aligned} f(n) &=t_{0} n^{\log _{2} k} \\ \text { where } f(n) &=\text { Time required to produce } n^{\text {th }} \text { unit } \end{aligned} $$ \(t_{0}=\) Time required to produce the first unit. The value of \(k\) will vary from industry to industry, but for a particular industry it is often a constant. (a) Suppose for a particular industry, the first unit takes 10,000 hours of labor to produce, so \(n_{0}=\) \(10,000,\) and that \(k=0.8\). The learning-curve effect states that the additional hours of labor goes down by a fixed percentage each time production doubles. Show that this is the case by completing the table. By what percent does the required time drop when production is doubled? $$ \begin{array}{c|c|c|c|c|c} \hline n & 1 & 2 & 4 & 8 & 16 \\ \hline f(n) & & & & & \\ \hline \end{array} $$ (b) What kind of function (exponential, power, logarithmic, other) is \(f(n) ?\) Discuss. (c) Suppose a different industry has a lower value of \(t_{0}\), say, \(t_{0}=7000\), but a higher value of \(b\), say, \(b=0.9 .\) Explain what this tells you about the difference in the learning-curve effect between these two industries. (d) What would it mean for \(k>1\) in terms of production time? Explain why this is unlikely to be the case.

Solve the equations, first approximately, as in Example \(1,\) by filling in the given table, and th猫n to four decimal places by using logarithms. $$ \begin{aligned} &\text { Table } 11.3 \text { Solve } 10^{x}=500\\\ &\begin{array}{c|c|c|c|c} \hline x & 2.6 & 2.7 & 2.8 & 2.9 \\ \hline 10^{x} & & & & \\ \hline \end{array} \end{aligned} $$

Rewrite the expression in terms of \(\log A\) and \(\log B\), or state that this is not possible. $$ \log \left(A B^{2}\right) $$

Rewrite the expression in terms of \(\log A\) and \(\log B\), or state that this is not possible. $$ \log \left(A B^{2}+B\right) $$

Without using a calculator, find two consecutive integers, one lying above and the other lying below the logarithm of the number. $$ 1.22 \cdot 10^{4} $$
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