Solve For { X $} . . . { 5^{2x-6} = 8 \}

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Introduction

Solving exponential equations can be a challenging task, but with the right approach, it can be done efficiently. In this article, we will focus on solving the equation $5^{2x-6} = 8$ for the variable $x$. This equation involves an exponential term with a base of 5 and an exponent that is a linear expression in terms of $x$. Our goal is to isolate the variable $x$ and find its value.

Understanding Exponential Equations

Exponential equations are equations that involve an exponential term, which is a number raised to a power. In this case, the exponential term is $5^{2x-6}$. The base of the exponential term is 5, and the exponent is $2x-6$. The equation states that this exponential term is equal to 8.

Properties of Exponential Equations

Exponential equations have several properties that can be used to solve them. One of the most important properties is the fact that if $a^x = a^y$, then $x = y$. This property can be used to eliminate the exponential term and solve for the variable.

Solving the Equation

To solve the equation $5^{2x-6} = 8$, we can start by taking the logarithm of both sides of the equation. This will allow us to eliminate the exponential term and solve for the variable.

Taking the Logarithm

Taking the logarithm of both sides of the equation gives us:

$$\log(5^{2x-6}) = \log(8)$$

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

$$(2x-6)\log(5) = \log(8)$$

Solving for $x$

Now that we have eliminated the exponential term, we can solve for the variable $x$. To do this, we can start by isolating the term $2x-6$ on one side of the equation.

$$2x-6 = \frac{\log(8)}{\log(5)}$$

Next, we can add 6 to both sides of the equation to get:

$$2x = \frac{\log(8)}{\log(5)} + 6$$

Finally, we can divide both sides of the equation by 2 to solve for $x$:

$$x = \frac{\frac{\log(8)}{\log(5)} + 6}{2}$$

Simplifying the Expression

The expression for $x$ can be simplified by evaluating the logarithms and performing the arithmetic operations.

$$x = \frac{\log(8) + 6\log(5)}{2\log(5)}$$

Using a calculator to evaluate the logarithms, we get:

$$x = \frac{0.90309 + 6(0.69897)}{2(0.69897)}$$

Simplifying the expression further, we get:

$$x = \frac{0.90309 + 4.19382}{1.39794}$$

$$x = \frac{5.09691}{1.39794}$$

$$x = 3.653$$

Conclusion

In this article, we solved the equation $5^{2x-6} = 8$ for the variable $x$. We used the properties of exponential equations and logarithms to eliminate the exponential term and solve for the variable. The final solution for $x$ is $x = 3.653$. This solution can be verified by plugging it back into the original equation to check that it is true.

Additional Tips and Tricks

• When solving exponential equations, it is often helpful to take the logarithm of both sides of the equation to eliminate the exponential term.
• Use the properties of logarithms to simplify the equation and make it easier to solve.
• Be careful when evaluating logarithms and performing arithmetic operations to avoid errors.
• Use a calculator to evaluate logarithms and perform arithmetic operations when necessary.

Final Thoughts

Solving exponential equations can be a challenging task, but with the right approach, it can be done efficiently. By using the properties of exponential equations and logarithms, we can eliminate the exponential term and solve for the variable. The final solution for $x$ is $x = 3.653$. This solution can be verified by plugging it back into the original equation to check that it is true.

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Introduction

In our previous article, we solved the equation $5^{2x-6} = 8$ for the variable $x$. We used the properties of exponential equations and logarithms to eliminate the exponential term and solve for the variable. In this article, we will answer some common questions that readers may have about solving exponential equations.

Q&A

Q: What is the first step in solving an exponential equation?

A: The first step in solving an exponential equation is to take the logarithm of both sides of the equation. This will allow us to eliminate the exponential term and solve for the variable.

Q: What is the property of logarithms that we use to eliminate the exponential term?

A: The property of logarithms that we use to eliminate the exponential term is $\log(a^b) = b\log(a)$. This property allows us to rewrite the exponential term as a linear expression in terms of the logarithm.

Q: How do we solve for the variable $x$ after taking the logarithm of both sides of the equation?

A: After taking the logarithm of both sides of the equation, we can solve for the variable $x$ by isolating the term $2x-6$ on one side of the equation. We can then add 6 to both sides of the equation and divide both sides of the equation by 2 to solve for $x$.

Q: What is the final solution for $x$ in the equation $5^{2x-6} = 8$?

A: The final solution for $x$ in the equation $5^{2x-6} = 8$ is $x = 3.653$. This solution can be verified by plugging it back into the original equation to check that it is true.

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

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

• Not taking the logarithm of both sides of the equation
• Not using the properties of logarithms to simplify the equation
• Not isolating the term $2x-6$ on one side of the equation
• Not adding 6 to both sides of the equation and dividing both sides of the equation by 2 to solve for $x$

Q: How can we verify the solution for $x$?

A: We can verify the solution for $x$ by plugging it back into the original equation to check that it is true. If the solution satisfies the original equation, then it is the correct solution.

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

A: Exponential equations have many real-world applications, including:

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

Conclusion

In this article, we answered some common questions that readers may have about solving exponential equations. We discussed the first step in solving an exponential equation, the property of logarithms that we use to eliminate the exponential term, and how to solve for the variable $x$ after taking the logarithm of both sides of the equation. We also discussed some common mistakes to avoid when solving exponential equations and how to verify the solution for $x$. Finally, we discussed some real-world applications of exponential equations.

Additional Tips and Tricks

• When solving exponential equations, it is often helpful to take the logarithm of both sides of the equation to eliminate the exponential term.
• Use the properties of logarithms to simplify the equation and make it easier to solve.
• Be careful when evaluating logarithms and performing arithmetic operations to avoid errors.
• Use a calculator to evaluate logarithms and perform arithmetic operations when necessary.
• Verify the solution for $x$ by plugging it back into the original equation to check that it is true.

Final Thoughts

Solving exponential equations can be a challenging task, but with the right approach, it can be done efficiently. By using the properties of exponential equations and logarithms, we can eliminate the exponential term and solve for the variable. The final solution for $x$ is $x = 3.653$. This solution can be verified by plugging it back into the original equation to check that it is true.