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Solve each equation. Express all solutions in exact form. $$5\left(10^{3 x}\right)-4=6$$

Short Answer

Expert verified
\(x = \frac{\log(2)}{3}\)

Step by step solution

01

Isolate the exponential term

Start by adding 4 to both sides of the equation to isolate the term involving the exponent. This gives us: \[ 5(10^{3x}) = 10 \]
02

Solve for the exponential term

Now, divide both sides of the equation by 5 to fully isolate the exponential term: \[ 10^{3x} = 2 \]
03

Use logarithms to solve for x

To solve for \(x\), take the logarithm (base 10) of both sides of the equation: \[ \log(10^{3x}) = \log(2) \] Using the logarithm property \( \log(a^b) = b \cdot \log(a) \), we get: \[ 3x \cdot \log(10) = \log(2) \] Since \(\log(10) = 1\), the equation simplifies to: \[ 3x = \log(2) \]
04

Solve for x

Finally, divide both sides of the equation by 3 to solve for \(x\): \[ x = \frac{\log(2)}{3} \] This is the exact form of the solution.

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Key Concepts

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

Logarithms
Logarithms are very useful when working with exponential equations, as they allow us to solve for unknown exponents. They are the inverse of exponential functions. For instance, if we have an equation like \( a^b = c \), we can use logarithms to solve for \( b \). This is because a logarithm answers the question: "To what power must the base be raised, to obtain a given number?"

Some key properties of logarithms include:
  • Logarithm of a Product: \( \log(ab) = \log(a) + \log(b) \) - This property allows us to break down the logarithm of a product into the sum of the logs.
  • Logarithm of a Quotient: \( \log(\frac{a}{b}) = \log(a) - \log(b) \) - This property is useful when dealing with division inside a log.
  • Logarithm of a Power: \( \log(a^b) = b \cdot \log(a) \) - This property is crucial for solving equations with exponents, as it allows the exponent to come out in front as a multiplier.
In our original problem, once we isolated the exponential expression, we took the logarithm of both sides. This step was necessary to "bring down" the exponent, allowing us to solve for \( x \). This manipulation showcases how powerful logarithms are in simplifying and solving equations.
Equation Solving
Solving equations involves finding values of the unknown variable that make the equation true. It is a fundamental part of algebra and ensures that we understand how changes in one part of the equation affect others.

There are various strategies for solving equations:
  • Isolation of variables: Move terms with the unknown variable to one side of the equation and constants to the other. This often involves using inverse operations, like addition if something was subtracted.
  • Balancing the equation: Whatever you do to one side of the equation, do the same to the other. This keeps the equation equal.
In our specific exercise, the aim was to isolate the term \( 10^{3x} \). Initially, we had a more complex expression, but by carefully balancing the equation through addition and division, we could simplify it to a basic form. This step-by-step isolation is crucial because it makes the effects of each operation clear, working our way carefully towards solving for \( x \).
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying equations to solve for an unknown variable. It involves performing legal mathematical operations to transform equations into a solvable form.

Key strategies in algebraic manipulation include:
  • Combining like terms: Gather similar terms (those involving the same variables) to simplify equations.
  • Using properties of equality: Understand and apply properties such as the distributive property (\( a(b + c) = ab + ac \)), associativity, and commutativity.
  • Inversion operations: Additive inverses cancel each other out (e.g., \( a - a = 0 \)) and multiplicative inverses (e.g., \( a \times \frac{1}{a} = 1 \)), which help in isolating variables.
In the solved problem, we used division to isolate the exponential term, which is an example of algebraic manipulation. This step reduced the equation to a simpler form that could be easily tackled with logarithms. Understanding these techniques helps not only with solving equations but also enhances overall algebraic prowess, making it easier to understand more complex mathematical concepts in the future.

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Most popular questions from this chapter

Suppose that when a ball is dropped, the height of its first rebound is about \(80 \%\) of the initial height that it was dropped from, the second rebound is about \(80 \%\) as high as the first rebound, and so on. If this ball is dropped from 12 feet in the air, model the height in feet of each rebound with an exponential function \(H(x),\) where \(x=0\) represents the initial height, \(x=1\) represents the height on the first rebound, and so on. Find the height of the third rebound. Determine which rebound had a height of about 2.5 feet.

In \(2012,63 \%\) of the U.S. population was non-Hispanic white, and this number is expected to be \(43 \%\) in \(2060 .\) (Source: U.S. Census Bureau.) (a) Find \(C\) and \(a\) so that \(P(x)=C a^{x-2012}\) models these data, where \(P\) is the percent of the population that is non-Hispanic white and \(x\) is the year. Why is \(a<1 ?\) (b) Estimate \(P\) in 2020 (c) Use \(P\) to estimate when \(50 \%\) of the population could be non-Hispanic white.

Use the properties of logarithms to rewrite each expression as a single logarithm with coefficient 1. Assume that all variables represent positive real numbers. $$\frac{4}{3} \ln m-\frac{2}{3} \ln 8 n-\ln m^{3} n^{2}$$

Traffic Flow \(\quad\) At an intersection, cars arrive randomly at an average rate of 30 cars per hour. Using the function $$ f(x)=1-e^{-0.5 x} $$ highway engineers estimate the likelihood or probability that at least one car will enter the intersection within a period of \(x\) minutes. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.) (a) Evaluate \(f(2)\) and interpret the answer. (b) Graph \(f\) for \(0 \leq x \leq 60 .\) What happens to the likelihood that at least one car enters the intersection during a 60 -minute period?

In 1666 the village of Eyam, located in England, experienced an outbreak of the Great Plague. Out of 261 people in the community, only 83 survived. The table shows a function \(f\) that computes the number of people who had not (yet) been infected after \(x\) days. $$\begin{array}{|c|r|r|r|r|}\hline x & 0 & 15 & 30 & 45 \\\\\hline f(x) & 254 & 240 & 204 & 150 \\\\\hline\end{array}$$ $$\begin{array}{|c|c|c|c|c|}\hline x & 60 & 75 & 90 & 125 \\\\\hline f(x) & 125 & 103 & 97 & 83\\\\\hline\end{array}$$ (a) Use a table to represent a function \(g\) that computes the number of people in Eyam who were infected after \(x\) days. (b) Write an equation that shows the relationship between \(f(x)\) and \(g(x)\) (c) Use graphing to decide which equation represents \(g(x)\) better \(y_{1}=\frac{171}{1+18.6 e^{-0.0747 x}}\) or \(y_{2}=18.3(1.024)^{x}\) (d) Use your results from parts (b) and (c) to find a formula for \(f(x)\)

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