Solve The Equation: $\left(2^x-1\right)(2x-1)=0$

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Introduction

In mathematics, solving equations is a fundamental concept that helps us find the value of unknown variables. Equations can be linear or non-linear, and they can involve various mathematical operations such as addition, subtraction, multiplication, and division. In this article, we will focus on solving a non-linear equation involving exponential and linear terms. The equation we will be solving is $\left(2^x-1\right)(2x-1)=0$. This equation involves a product of two terms, and we need to find the values of $x$ that make the equation true.

Understanding the Equation

The given equation is a product of two terms: $\left(2^x-1\right)$ and $(2x-1)$. For the product of two terms to be zero, at least one of the terms must be zero. Therefore, we can set each term equal to zero and solve for $x$. This will give us the values of $x$ that satisfy the equation.

Solving the First Term

Let's start by solving the first term: $\left(2^x-1\right)=0$. To solve this equation, we can isolate the exponential term by adding $1$ to both sides of the equation. This gives us:

$$2^x=1$$

Now, we can take the logarithm of both sides of the equation to solve for $x$. We can use any base for the logarithm, but let's use the natural logarithm (base $e$) for simplicity. Taking the natural logarithm of both sides gives us:

$$\ln(2^x)=\ln(1)$$

Using the property of logarithms that states $\ln(a^b)=b\ln(a)$, we can simplify the left-hand side of the equation:

$$x\ln(2)=0$$

Now, we can divide both sides of the equation by $\ln(2)$ to solve for $x$:

$$x=0$$

Solving the Second Term

Now, let's solve the second term: $(2x-1)=0$. To solve this equation, we can add $1$ to both sides of the equation and then divide both sides by $2$. This gives us:

$$2x=1$$

Dividing both sides of the equation by $2$ gives us:

$$x=\frac{1}{2}$$

Combining the Solutions

We have found two solutions for the equation: $x=0$ and $x=\frac{1}{2}$. However, we need to check if these solutions satisfy the original equation. Substituting $x=0$ into the original equation gives us:

$$\left(2^0-1\right)(2(0)-1)=\left(1-1\right)(-1)=0$$

This shows that $x=0$ is indeed a solution to the equation. Substituting $x=\frac{1}{2}$ into the original equation gives us:

$$\left(2^{\frac{1}{2}}-1\right)(2\left(\frac{1}{2}\right)-1)=\left(\sqrt{2}-1\right)(1-1)=0$$

This shows that $x=\frac{1}{2}$ is also a solution to the equation.

Conclusion

In this article, we solved the equation $\left(2^x-1\right)(2x-1)=0$ by setting each term equal to zero and solving for $x$. We found two solutions: $x=0$ and $x=\frac{1}{2}$. We then checked if these solutions satisfied the original equation, and we found that both solutions are indeed valid. This shows that the equation has two distinct solutions, and we can use these solutions to model real-world problems involving exponential and linear terms.

Applications of the Equation

The equation $\left(2^x-1\right)(2x-1)=0$ has several applications in mathematics and science. For example, it can be used to model population growth, where the exponential term represents the growth rate and the linear term represents the initial population. It can also be used to model chemical reactions, where the exponential term represents the concentration of a reactant and the linear term represents the rate of reaction.

Final Thoughts

Solving equations is a fundamental concept in mathematics, and it has numerous applications in science and engineering. In this article, we solved the equation $\left(2^x-1\right)(2x-1)=0$ by setting each term equal to zero and solving for $x$. We found two solutions: $x=0$ and $x=\frac{1}{2}$. We then checked if these solutions satisfied the original equation, and we found that both solutions are indeed valid. This shows that the equation has two distinct solutions, and we can use these solutions to model real-world problems involving exponential and linear terms.

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "Calculus" by Michael Spivak
• [3] "Mathematics for Computer Science" by Eric Lehman and Tom Leighton

Further Reading

• [1] "Solving Equations" by Khan Academy
• [2] "Exponential and Logarithmic Functions" by Mathway
• [3] "Linear and Quadratic Equations" by Purplemath
  

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Introduction

In our previous article, we solved the equation $\left(2^x-1\right)(2x-1)=0$ by setting each term equal to zero and solving for $x$. We found two solutions: $x=0$ and $x=\frac{1}{2}$. In this article, we will answer some frequently asked questions about solving this equation.

Q: What is the significance of the equation $\left(2^x-1\right)(2x-1)=0$?

A: The equation $\left(2^x-1\right)(2x-1)=0$ is significant because it involves a product of two terms, and we need to find the values of $x$ that make the equation true. This equation has applications in mathematics and science, such as modeling population growth and chemical reactions.

Q: How do I solve the equation $\left(2^x-1\right)(2x-1)=0$?

A: To solve the equation $\left(2^x-1\right)(2x-1)=0$, we need to set each term equal to zero and solve for $x$. This involves taking the logarithm of both sides of the equation to solve for $x$.

Q: What are the solutions to the equation $\left(2^x-1\right)(2x-1)=0$?

A: The solutions to the equation $\left(2^x-1\right)(2x-1)=0$ are $x=0$ and $x=\frac{1}{2}$. These solutions satisfy the original equation and can be used to model real-world problems.

Q: How do I check if the solutions satisfy the original equation?

A: To check if the solutions satisfy the original equation, we need to substitute the solutions into the original equation and verify that the equation is true.

Q: What are some applications of the equation $\left(2^x-1\right)(2x-1)=0$?

A: The equation $\left(2^x-1\right)(2x-1)=0$ has applications in mathematics and science, such as modeling population growth and chemical reactions.

Q: Can I use the equation $\left(2^x-1\right)(2x-1)=0$ to model real-world problems?

A: Yes, the equation $\left(2^x-1\right)(2x-1)=0$ can be used to model real-world problems involving exponential and linear terms.

Q: How do I use the equation $\left(2^x-1\right)(2x-1)=0$ to model population growth?

A: To use the equation $\left(2^x-1\right)(2x-1)=0$ to model population growth, we need to substitute the growth rate into the exponential term and the initial population into the linear term.

Q: How do I use the equation $\left(2^x-1\right)(2x-1)=0$ to model chemical reactions?

A: To use the equation $\left(2^x-1\right)(2x-1)=0$ to model chemical reactions, we need to substitute the concentration of a reactant into the exponential term and the rate of reaction into the linear term.

Conclusion

In this article, we answered some frequently asked questions about solving the equation $\left(2^x-1\right)(2x-1)=0$. We discussed the significance of the equation, how to solve it, and its applications in mathematics and science. We also provided examples of how to use the equation to model real-world problems.

References

• [1] "Algebra and Trigonometry" by Michael Sullivan
• [2] "Calculus" by Michael Spivak
• [3] "Mathematics for Computer Science" by Eric Lehman and Tom Leighton

Further Reading

• [1] "Solving Equations" by Khan Academy
• [2] "Exponential and Logarithmic Functions" by Mathway
• [3] "Linear and Quadratic Equations" by Purplemath