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Chapter 2: Problem 7

Solve the following equations for \(x\) : (a) \(3^{x}=7\) (b) \(5(2)^{x}=10^{x}\)

Short Answer

Expert verified
For the first equation, \( x \approx 1.7712 \). For the second equation, \( x = 1 \).

Step by step solution

01

Solve for x in the first equation \( 3^{x}=7 \)

To solve the exponential equation \( 3^{x}=7 \), take the natural logarithm of both sides: \( \text{ln}(3^{x}) = \text{ln}(7) \). Using the logarithm power rule \( \text{ln}(a^b) = b \text{ln}(a) \), rewrite the equation as \( x \text{ln}(3) = \text{ln}(7) \). Divide both sides by \( \text{ln}(3) \): \( x = \frac{\text{ln}(7)}{\text{ln}(3)} \). Calculate the value: \( x \approx 1.7712 \).
02

Solve for x in the second equation \( 5(2)^{x}=10^{x} \)

Start by rewriting the equation in terms of the same base. Note that \( 10 = 2 \times 5 \). So, express \( 10^{x} \) as \( (2 \times 5)^{x} \). Rewrite the equation: \( 5(2)^{x} = (2 \times 5)^{x} \). Apply the property of exponents: \( 5(2)^{x} = 2^{x} \times 5^{x} \). Divide both sides by \( 2^{x} \) to simplify the equation: \( 5 = 5^{x} \). Solve for \( x \) by recognizing that both sides must be equal: \( x = 1 \).

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Key Concepts

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

natural logarithms
Natural logarithms are a type of logarithm with the base of the natural number e (approximately 2.71828). They are often denoted as ln. Natural logarithms are useful when dealing with exponential equations of the form \( a^x = b \). By taking the natural logarithm of both sides, we can easily solve for x. For example, in the equation \( 3^x = 7 \), taking the natural logarithm of both sides gives us \( \text{ln}(3^x) = \text{ln}(7) \). Using logarithms simplifies complex calculations, making them easier and more manageable.
logarithm power rule
The logarithm power rule is a fundamental property that states \( \text{log}_a(b^c) = c \cdot \text{log}_a(b) \). This rule allows us to move the exponent in a logarithmic expression to a multiplier of the logarithm. For example, in solving \( 3^x = 7 \), after taking the natural logarithm of both sides, we use the power rule: \( \text{ln}(3^x) = x \cdot \text{ln}(3) \). By expressing the equation this way, we can isolate the variable x and solve for it effectively.
exponential equations
Exponential equations are equations in which variables appear as exponents. A common method to solve exponential equations is to use logarithms. When faced with \( 3^x = 7 \), one intuitive approach is to apply the natural logarithm to both sides of the equation. By leveraging properties of logarithms and exponents, we can transform and solve the equation. Additionally, when equations involve expressions like \( 10^x = 2 \cdot 5^x \), recognizing how to rewrite bases and apply rules of exponents helps break down and solve complex expressions.
properties of exponents
Properties of exponents refer to a set of rules that govern operations on exponential expressions. Key properties include:
  • Product of powers: \( a^m \cdot a^n = a^{m+n} \)
  • Quotient of powers: \( \frac{a^m}{a^n} = a^{m-n} \)
  • Power of a power: \( (a^m)^n = a^{m \cdot n} \)
  • Power of a product: \( (ab)^m = a^m \cdot b^m \)
In the equation \( 5(2^x) = (2 \cdot 5)^x \), we can use these properties to rewrite and simplify the expressions. Recognizing that \( (2 \cdot 5)^x = 2^x \cdot 5^x \) allows us to cancel terms efficiently and solve for the variable x.

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Most popular questions from this chapter

The number of items, \(N\), produced each day by an assembly-line worker, \(t\) days after an initial training period, is modelled by $$ N=100-100 \mathrm{e}^{-0.4 t} $$ (1) Calculate the number of items produced daily (a) 1 day after the training period (b) 2 days after the training period (c) 10 days after the training period (2) What is the worker's daily production in the long run? (3) Sketch a graph of \(N\) against \(t\) and explain why the general shape might have been expected.

(Excel) The demand function of a good can be modelled approximately by $$ P=100-\frac{2}{3} Q^{n} $$ (a) Show that if this relation is exact then a graph of \(\ln (150-1.5 P)\) against In \(Q\) will be a straight line passing through the origin with slope \(n\). (b) For the data given below, tabulate the values of \(\ln (150-1.5 P)\) and \(\ln Q\). Find the line of best fit and hence estimate the value of \(n\) correct to 1 decimal place. \begin{tabular}{lllllll} \(Q\) & 10 & 50 & 60 & 100 & 200 & 400 \\ \hline\(P\) & 95 & 85 & 80 & 70 & 50 & 20 \end{tabular}

Given that fixed costs are 100 and that variable costs are 2 per unit, express \(\mathrm{TC}\) and \(\mathrm{AC}\) as functions of \(Q\). Hence sketch their graphs.

Solve the equation \(f(x)=0\) for each of the following quadratic functions: (a) \(f(x)=x^{2}-16\) (b) \(f(x)=x(100-x)\) (c) \(f(x)=-x^{2}+22 x-85\) (d) \(f(x)=x^{2}-18 x+81\) (e) \(f(x)=2 x^{2}+4 x+3\)

Given the quadratic supply and demand functions $$ \begin{aligned} &P=Q_{\mathrm{s}}^{2}+2 Q_{\mathrm{s}}+12 \\ &P=-Q_{\mathrm{D}}^{2}-4 Q_{\mathrm{D}}+68 \end{aligned} $$ determine the equilibrium price and quantity.
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