/*! 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 34 Solve the equation for \(x\). ... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the equation for \(x\). \(3 \cdot 10^{2-5 x}=4\)

Short Answer

Expert verified
The solution is \( x = \frac{2 - \log_{10}\left(\frac{4}{3}\right)}{5} \).

Step by step solution

01

Simplify Exponential Equation

Start with the given equation: \[ 3 \cdot 10^{2 - 5x} = 4 \] Divide both sides by 3 to leave the exponential term by itself:\[ 10^{2 - 5x} = \frac{4}{3} \]
02

Take the Logarithm of Both Sides

To solve for the exponent, take the logarithm base 10 of both sides:\[ \log_{10}(10^{2 - 5x}) = \log_{10}\left(\frac{4}{3}\right) \] Applying the properties of logarithms, this simplifies to:\[ 2 - 5x = \log_{10}\left(\frac{4}{3}\right) \]
03

Solve for x

Rearrange the equation to solve for \(x\):\[ 2 - 5x = \log_{10}\left(\frac{4}{3}\right) \] Subtract 2 from both sides:\[ -5x = \log_{10}\left(\frac{4}{3}\right) - 2 \] Divide both sides by -5:\[ x = \frac{2 - \log_{10}\left(\frac{4}{3}\right)}{5} \]

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Key Concepts

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

Logarithmic Functions
Logarithmic functions are the inverses of exponential functions. When you need to find the unknown exponent in an equation like the one in our problem, logarithms act as a very helpful tool.
Logarithmic functions are denoted as \(\log_b(a)\), which means "the power to which we must raise the base \(b\) to get \(a\)."
For instance, \(\log_{10}(100) = 2\) because 10 raised to the power of 2 is 100.
  • **Inverse Relationship**: The key role of logarithms is being the inverse of exponentiation. This means they can reverse the process of raising a base to a power.
  • **Common Bases**: The most commonly used bases are 10 and \(e\), known as common and natural logarithms, respectively.
Using this function allows us to bring the exponent down, out of its base, and simplifies solving equations where the variable is in the exponent. In the problem you are dealing with, by taking the logarithm of both sides, we were able to transform the exponential equation into a manageable form that could be easily solved for \(x\).
Properties of Logarithms
Logarithms come with a set of properties that make manipulating exponential equations easier. These properties enable us to simplify complex expressions and solve equations effectively.
One fundamental property is:
  • **Power Rule**: This rule states that \(\log_b(m^n) = n \log_b(m)\). It's particularly useful for solving exponential equations because it allows us to bring exponents down as coefficients.
For example, in our exercise, \(\log_{10}(10^{2-5x})\), using the power rule, changes into \(2-5x\cdot\log_{10}(10)\). Since \(\log_{10}(10) = 1\), it's straightforward to simplify further.
There are other handy properties to remember:
  • **Product Rule**: \(\log_b(mn) = \log_b(m) + \log_b(n)\)
  • **Quotient Rule**: \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\)
These properties enable us to break down or combine logarithmic expressions to aid in solving equations, making logarithms very versatile in algebra.
Exponential Functions
Exponential functions are mathematical expressions where the variable appears as an exponent. They take the form \(f(x) = a \cdot b^x\), where \(a\) is a constant multiplier, and \(b\) is the base raised to the power of \(x\).
In our exercise, the equation \(3 \cdot 10^{2-5x}\) is an example of an exponential function.Understanding exponential functions is crucial:
  • **Growth and Decay**: Exponential functions model growth or decay processes such as populations, radioactive decay, and interest calculations. Depending on the base \(b\), they show different behaviors. When \(b>1\), the function indicates exponential growth. Conversely, if \(0
  • **Rapid Change**: The rate of change in exponential growth is faster compared to linear growth, which is why these functions are compelling when modeling real-life phenomenons.
When solving exponential equations, like the one in this exercise, the goal is usually to isolate the exponential part and then use logarithms to find the unknown variable, as we have done in the steps provided. This approach uncovers the power tied up in exponential expressions, showing the crucial role exponential functions play in mathematics.

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Most popular questions from this chapter

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