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Chapter 10: Problem 56

Solve the equations in Problem for \(x\). Your solutions will involve \(u\). \(8^{x}=2^{u}\)

Short Answer

Expert verified
Question: Given the equation \(8^{x}=2^{u}\), find the value of \(x\) in terms of \(u\). Answer: \(x=\frac{u}{3}\).

Step by step solution

01

Rewrite the base 8 as 2

We can rewrite the base \(8\) by expressing it as a power of \(2\). Since \(8=2^{3}\), we can substitute \(8\) by \(2^{3}\). So, our equation becomes: \((2^{3})^{x}=2^{u}\).
02

Simplify the left side of the equation

Now, we should simplify the left side of the equation using the rule for exponents which states \((a^{m})^{n}=a^{mn}\): \(2^{3x}=2^{u}\).
03

Equate the exponents

Since the equation \(2^{3x}=2^{u}\) has the same base on both sides, we can equate their exponents. That is: \(3x=u\).
04

Solve for \(x\)

To solve for \(x\), divide both sides of the equation by \(3\): \(x=\frac{u}{3}\). Now we have found the solution for \(x\) in terms of \(u\).

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Key Concepts

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

Exponent Rules
Exponent rules are like shortcuts for working with powers of numbers. They help us simplify expressions that involve exponents. Here are a few key rules:
  • Product of Powers: When you multiply two expressions with the same base, such as \( a^m \times a^n \), you add the exponents. This results in \( a^{m+n} \).
  • Power of a Power: If you have an exponent raised to another exponent, like \((a^m)^n\), you multiply the exponents. This will simplify to \( a^{mn} \).
  • Base Conversion: Sometimes, it's helpful to express numbers as powers of another smaller number, like expressing \(8\) as \(2^3\).
In the exercise, these rules help us rewrite \(8^x\) as \((2^3)^x\), and then simplify it to \(2^{3x}\). Recognizing the same base allows us to equate their exponents.
Solving Equations
Solving equations involves finding the value(s) of the variable that make the equation true. In algebra, this often means "isolating" the variable on one side of the equation. The steps generally include:
  • Rewriting the equation to a simpler form, often achieved by eliminating parentheses or combining like terms.
  • Moving all instances of the variable to one side of the equation and constant terms to the other using addition, subtraction, multiplication, or division.
  • Using inverse operations to solve for the variable.
When solving \(8^x = 2^u\), the equation was first rewritten using powers of 2, transformed into \( (2^3)^x = 2^u \). The simplified equation was \( 2^{3x} = 2^u \), and we used the fact that if the bases are equal, then the exponents must be equal.
Variable Manipulation
Variable manipulation involves changing the form of an equation to allow easier solving or substituting into another equation. Here鈥檚 how it works:
  • Identify parts of the equation that involve variables and look for simplifications.
  • If an expression relates to a variable, manipulate it to make solving straightforward. For instance, expressing a complex number relationship in terms of one variable.
  • Use rules and operations actively to transform expressions, such as isolating a variable by dividing or multiplying by a constant.
In the step-by-step solution, the method involved equating exponents since the bases were identical, leading from \(3x = u\) to \(x = \frac{u}{3}\). Solving for \(x\) involved dividing both sides of the equation by 3, resulting in an expression that describes \(x\) in terms of \(u\). Mastering these skills allows for flexibility and precision in solving algebraic equations.

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Most popular questions from this chapter

(a) Construct a table of values for the function \(g(x)=\) \(28(1.1)^{x}\) for \(x=0,1,2,3,4\) (b) For which values of \(x\) in the table is (i) \(g(x)<33.88\) (ii) \(g(x)>30.8\) (iii) \(g(x)=37.268\)

State the starting value \(a\), the growth factor \(b\), and the growth rate \(r\) as a percent correct to 2 decimals for the exponential functions. . \(V=2000(1.0058)^{t}\)

The percent change (increase or decrease) in the value of an investment each year over a five-year period is shown in Table 10.7 . By what percent does the investment's value change over this five-year period? $$ \begin{array}{c|c|c|c|c|c} \hline t & 1 & 2 & 3 & 4 & 5 \\ \hline r & 12.9 \% & 9.2 \% & 11.3 \% & -4.5 \% & -13.6 \% \\ \hline \end{array} $$

Match each statement (a)-(b) with the solutions to one or more of the equations (I)-(VI). I. \(10(1.2)^{t}=5\) II. \(10=5(1.2)^{t}\) III. \(10+5(1.2)^{t}=0\) IV. \(5+10(1.2)^{t}=0\) V. \(10(0.8)^{t}=5\) VI. \(5(0.8)^{t}=10\) (a) The time an exponentially growing quantity takes to grow from 5 to 10 grams. (b) The time an exponentially decaying quantity takes to dron from 10 to 5 grams

Decide for what values of the constant \(A\) the equation has (a) A solution (b) The solution \(t=0\) (c) A positive solution \(6.3 A-3 \cdot 7^{t}=0\)
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