Solve The Equation:$\[ 25^{2x-1} = \frac{1}{5} X^{2x+1} \\]

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Introduction

Solving equations involving exponents can be a challenging task, especially when they involve variables in the exponent. In this article, we will focus on solving the equation $25^{2x-1} = \frac{1}{5} x^{2x+1}$, which involves a base of 25 and a variable in the exponent. We will use various mathematical techniques to simplify the equation and solve for the variable x.

Understanding the Equation

The given equation is $25^{2x-1} = \frac{1}{5} x^{2x+1}$. To start solving this equation, we need to understand the properties of exponents and how to manipulate them. The base of the exponent is 25, which can be expressed as $5^2$. We can rewrite the equation as $(5^2)^{2x-1} = \frac{1}{5} x^{2x+1}$.

Simplifying the Equation

Using the property of exponents that $(a^m)^n = a^{mn}$, we can simplify the left-hand side of the equation:

$$(5^2)^{2x-1} = 5^{2(2x-1)} = 5^{4x-2}$$

Now, the equation becomes:

$$5^{4x-2} = \frac{1}{5} x^{2x+1}$$

Manipulating the Equation

To make the equation easier to solve, we can multiply both sides by $5$ to get rid of the fraction:

$$5 \cdot 5^{4x-2} = x^{2x+1}$$

Using the property of exponents that $a^m \cdot a^n = a^{m+n}$, we can simplify the left-hand side of the equation:

$$5^{4x-1} = x^{2x+1}$$

Using Logarithms to Solve the Equation

Since the equation involves a variable in the exponent, we can use logarithms to solve for x. Taking the logarithm of both sides of the equation, we get:

$$\log(5^{4x-1}) = \log(x^{2x+1})$$

Using the property of logarithms that $\log(a^b) = b \log(a)$, we can simplify the equation:

$$(4x-1) \log(5) = (2x+1) \log(x)$$

Solving for x

Now, we can solve for x by isolating the variable on one side of the equation. We can start by multiplying both sides by $\frac{1}{\log(5)}$ to get rid of the logarithm on the left-hand side:

$$4x-1 = \frac{2x+1}{\log(5)} \log(x)$$

Multiplying both sides by $\log(5)$, we get:

$$(4x-1) \log(5) = (2x+1) \log(x) \log(5)$$

Expanding the right-hand side of the equation, we get:

$$(4x-1) \log(5) = 2x \log(x) \log(5) + \log(x) \log(5)$$

Subtracting $2x \log(x) \log(5)$ from both sides, we get:

$$(4x-1) \log(5) - 2x \log(x) \log(5) = \log(x) \log(5)$$

Factoring out $\log(5)$ from the left-hand side of the equation, we get:

$$\log(5) (4x-1 - 2x \log(x)) = \log(x) \log(5)$$

Dividing both sides by $\log(5)$, we get:

$$4x-1 - 2x \log(x) = \log(x)$$

Adding $2x \log(x)$ to both sides, we get:

$$4x-1 = 3x \log(x)$$

Subtracting $4x$ from both sides, we get:

$$-1 = 3x \log(x) - 4x$$

Adding $4x$ to both sides, we get:

$$3x \log(x) = 3x$$

Dividing both sides by $3x$, we get:

$$\log(x) = 1$$

Conclusion

In this article, we have solved the equation $25^{2x-1} = \frac{1}{5} x^{2x+1}$ using various mathematical techniques. We started by simplifying the equation using the properties of exponents and then used logarithms to solve for x. The final solution is $\log(x) = 1$, which implies that $x = e^1 = e$.

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Introduction

In our previous article, we solved the equation $25^{2x-1} = \frac{1}{5} x^{2x+1}$ using various mathematical techniques. In this article, we will provide a Q&A section to help clarify any doubts or questions that readers may have.

Q: What is the base of the exponent in the equation?

A: The base of the exponent in the equation is 25, which can be expressed as $5^2$.

Q: How did you simplify the left-hand side of the equation?

A: We used the property of exponents that $(a^m)^n = a^{mn}$ to simplify the left-hand side of the equation. Specifically, we rewrote $(5^2)^{2x-1}$ as $5^{2(2x-1)} = 5^{4x-2}$.

Q: Why did you multiply both sides of the equation by 5?

A: We multiplied both sides of the equation by 5 to get rid of the fraction on the right-hand side of the equation. This made it easier to simplify the equation and solve for x.

Q: How did you use logarithms to solve the equation?

A: We took the logarithm of both sides of the equation to get rid of the exponent on the left-hand side. This allowed us to simplify the equation and solve for x.

Q: What is the final solution to the equation?

A: The final solution to the equation is $\log(x) = 1$, which implies that $x = e^1 = e$.

Q: Can you explain why we used logarithms to solve the equation?

A: We used logarithms to solve the equation because the equation involves a variable in the exponent. Logarithms allow us to get rid of the exponent and solve for the variable.

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

A: Some common mistakes to avoid when solving equations involving exponents include:

• Not using the correct properties of exponents
• Not simplifying the equation enough
• Not using logarithms when necessary
• Not checking the solution to make sure it satisfies the original equation

Q: How can I apply the techniques used in this article to solve other equations involving exponents?

A: The techniques used in this article can be applied to solve other equations involving exponents by following these steps:

• Simplify the equation using the properties of exponents
• Use logarithms to get rid of the exponent
• Solve for the variable
• Check the solution to make sure it satisfies the original equation

Conclusion

In this Q&A article, we have provided answers to common questions about solving the equation $25^{2x-1} = \frac{1}{5} x^{2x+1}$. We hope that this article has helped to clarify any doubts or questions that readers may have had.