Solve: E 10 X + E 5 X = 20 E^{10x} + E^{5x} = 20 E 10 X + E 5 X = 20 X = X = X = [Enter Your Answers, Separated By Commas; If No Solution Exists, Enter DNE]

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Introduction

Exponential equations can be challenging to solve, especially when they involve different bases and exponents. In this article, we will focus on solving the equation e10x+e5x=20e^{10x} + e^{5x} = 20, where ee is the base of the natural logarithm. We will use various techniques to isolate the variable xx and find its value.

Understanding Exponential Equations

Exponential equations are equations that involve an exponential function, which is a function of the form f(x)=axf(x) = a^x, where aa is a positive real number and xx is the variable. In the equation e10x+e5x=20e^{10x} + e^{5x} = 20, we have two exponential functions with different bases and exponents.

Properties of Exponential Functions

Before we can solve the equation, we need to understand some properties of exponential functions. One of the most important properties is the fact that exponential functions are one-to-one functions, which means that they pass the horizontal line test. This means that if f(x)=f(y)f(x) = f(y), then x=yx = y.

Using the Properties of Exponential Functions to Simplify the Equation

We can use the properties of exponential functions to simplify the equation e10x+e5x=20e^{10x} + e^{5x} = 20. Since the bases are the same, we can combine the two exponential functions using the rule ax+ay=ax+ya^x + a^y = a^{x+y}.

Applying the Rule to the Equation

Using the rule ax+ay=ax+ya^x + a^y = a^{x+y}, we can rewrite the equation as e10x+e5x=e5x(e5x+1)=20e^{10x} + e^{5x} = e^{5x}(e^{5x} + 1) = 20. This simplifies the equation and makes it easier to solve.

Isolating the Variable xx

Now that we have simplified the equation, we can isolate the variable xx. We can start by dividing both sides of the equation by e5xe^{5x}, which gives us e5x+1=20e5xe^{5x} + 1 = \frac{20}{e^{5x}}.

Using Algebraic Manipulation to Isolate xx

We can use algebraic manipulation to isolate xx. We can start by multiplying both sides of the equation by e5xe^{5x}, which gives us e10x+e5x=20e^{10x} + e^{5x} = 20. This is the same equation we started with, so we need to try a different approach.

Using the Natural Logarithm to Solve the Equation

One way to solve the equation is to use the natural logarithm. We can take the natural logarithm of both sides of the equation, which gives us ln(e10x+e5x)=ln(20)\ln(e^{10x} + e^{5x}) = \ln(20).

Using the Properties of the Natural Logarithm to Simplify the Equation

We can use the properties of the natural logarithm to simplify the equation. Since the natural logarithm is the inverse of the exponential function, we can rewrite the equation as 10x+5x=ln(20)10x + 5x = \ln(20).

Solving for xx

Now that we have simplified the equation, we can solve for xx. We can start by combining the two terms on the left-hand side of the equation, which gives us 15x=ln(20)15x = \ln(20).

Using Algebraic Manipulation to Solve for xx

We can use algebraic manipulation to solve for xx. We can start by dividing both sides of the equation by 15, which gives us x=ln(20)15x = \frac{\ln(20)}{15}.

Conclusion

In this article, we have solved the equation e10x+e5x=20e^{10x} + e^{5x} = 20 using various techniques. We have used the properties of exponential functions, the natural logarithm, and algebraic manipulation to isolate the variable xx and find its value. The solution to the equation is x=ln(20)15x = \frac{\ln(20)}{15}.

Final Answer

The final answer is ln(20)15\boxed{\frac{\ln(20)}{15}}.

Introduction

In our previous article, we solved the equation e10x+e5x=20e^{10x} + e^{5x} = 20 using various techniques. In this article, we will provide a Q&A guide to help you understand the solution and answer any questions you may have.

Q: What is the main concept behind solving exponential equations?

A: The main concept behind solving exponential equations is to use the properties of exponential functions, such as the rule ax+ay=ax+ya^x + a^y = a^{x+y}, and the natural logarithm to simplify the equation and isolate the variable.

Q: How do I simplify the equation e10x+e5x=20e^{10x} + e^{5x} = 20?

A: To simplify the equation, you can use the rule ax+ay=ax+ya^x + a^y = a^{x+y} to combine the two exponential functions. This will give you e5x(e5x+1)=20e^{5x}(e^{5x} + 1) = 20.

Q: How do I isolate the variable xx?

A: To isolate the variable xx, you can use algebraic manipulation to get the equation in the form x=ln(20)15x = \frac{\ln(20)}{15}.

Q: What is the final answer to the equation e10x+e5x=20e^{10x} + e^{5x} = 20?

A: The final answer to the equation is x=ln(20)15x = \frac{\ln(20)}{15}.

Q: Can I use other methods to solve the equation e10x+e5x=20e^{10x} + e^{5x} = 20?

A: Yes, you can use other methods to solve the equation, such as using the quadratic formula or graphing the equation. However, the method we used in this article is the most straightforward and efficient way to solve the equation.

Q: What are some common mistakes to avoid when solving exponential equations?

A: Some common mistakes to avoid when solving exponential equations include:

  • Not using the properties of exponential functions correctly
  • Not using the natural logarithm to simplify the equation
  • Not isolating the variable correctly
  • Not checking the solution for extraneous solutions

Q: How do I check the solution for extraneous solutions?

A: To check the solution for extraneous solutions, you can plug the solution back into the original equation and check if it is true. If the solution is not true, then it is an extraneous solution.

Q: What are some real-world applications of exponential equations?

A: Exponential equations have many real-world applications, such as:

  • Modeling population growth
  • Modeling chemical reactions
  • Modeling financial investments
  • Modeling electrical circuits

Conclusion

In this article, we have provided a Q&A guide to help you understand the solution to the equation e10x+e5x=20e^{10x} + e^{5x} = 20. We have also discussed some common mistakes to avoid when solving exponential equations and provided some real-world applications of exponential equations.

Final Answer

The final answer is ln(20)15\boxed{\frac{\ln(20)}{15}}.