Solve The Following Equation For X X X :i. 6 1 − X = 2 3 X + 1 6^{1-x} = 2^{3x+1} 6 1 − X = 2 3 X + 1

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Introduction

In this article, we will delve into solving the given equation for x, which involves exponential terms with different bases. The equation is $6^{1-x} = 2^{3x+1}$, and our goal is to isolate the variable x and find its value. We will use various mathematical techniques, including logarithmic properties and algebraic manipulations, to solve this equation.

Step 1: Take the Logarithm of Both Sides

To solve the equation, we can start by taking the logarithm of both sides. This will allow us to use the properties of logarithms to simplify the equation and isolate the variable x. We can use the natural logarithm (ln) or the common logarithm (log) as the base of the logarithm.

Taking the logarithm of both sides, we get:

$$\ln(6^{1-x}) = \ln(2^{3x+1})$$

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

$$(1-x)\ln(6) = (3x+1)\ln(2)$$

Step 2: Distribute the Logarithms

Next, we can distribute the logarithms on both sides of the equation:

$$\ln(6) - x\ln(6) = 3x\ln(2) + \ln(2)$$

Step 3: Isolate the Variable x

Now, we can isolate the variable x by moving all the terms involving x to one side of the equation:

$$-x\ln(6) - 3x\ln(2) = \ln(2) - \ln(6)$$

Step 4: Factor Out the Variable x

We can factor out the variable x from the left-hand side of the equation:

$$x(-\ln(6) - 3\ln(2)) = \ln(2) - \ln(6)$$

Step 5: Solve for x

Finally, we can solve for x by dividing both sides of the equation by the coefficient of x:

$$x = \frac{\ln(2) - \ln(6)}{-\ln(6) - 3\ln(2)}$$

Simplifying the Expression

We can simplify the expression by combining the logarithms in the numerator and denominator:

$$x = \frac{\ln\left(\frac{2}{6}\right)}{\ln\left(\frac{6}{2}\right) + 3\ln(2)}$$

$$x = \frac{\ln\left(\frac{1}{3}\right)}{\ln(3) + 3\ln(2)}$$

Conclusion

In this article, we have solved the equation $6^{1-x} = 2^{3x+1}$ for x using logarithmic properties and algebraic manipulations. The solution involves taking the logarithm of both sides, distributing the logarithms, isolating the variable x, factoring out the variable x, and solving for x. The final expression for x is a complex fraction involving logarithms.

Final Answer

The final answer is:

$$x = \frac{\ln\left(\frac{1}{3}\right)}{\ln(3) + 3\ln(2)}$$

This is the solution to the equation $6^{1-x} = 2^{3x+1}$ for x.

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Introduction

In our previous article, we solved the equation $6^{1-x} = 2^{3x+1}$ for x using logarithmic properties and algebraic manipulations. In this article, we will answer some common questions that readers may have about the solution.

Q: What is the significance of taking the logarithm of both sides of the equation?

A: Taking the logarithm of both sides of the equation is a common technique used to solve equations involving exponential terms. By taking the logarithm, we can use the properties of logarithms to simplify the equation and isolate the variable x.

Q: Why did we use the natural logarithm (ln) instead of the common logarithm (log)?

A: We used the natural logarithm (ln) because it is a more commonly used base in mathematics and is often easier to work with. However, the common logarithm (log) can also be used, and the solution would be similar.

Q: How did we simplify the expression $\ln\left(\frac{1}{3}\right)$?

A: We simplified the expression $\ln\left(\frac{1}{3}\right)$ by using the property of logarithms that states $\ln(a) = \ln(b) + \ln\left(\frac{a}{b}\right)$. In this case, we can rewrite $\frac{1}{3}$ as $\frac{1}{3} = \frac{1}{2} \cdot \frac{1}{3/2}$, and then take the logarithm of both sides.

Q: What is the relationship between the logarithm and the exponential function?

A: The logarithm and the exponential function are inverse functions. This means that if $y = \ln(x)$, then $x = e^y$, and vice versa. This relationship is useful in solving equations involving exponential terms.

Q: Can we use other methods to solve the equation $6^{1-x} = 2^{3x+1}$?

A: Yes, we can use other methods to solve the equation $6^{1-x} = 2^{3x+1}$. For example, we can use the change of base formula to rewrite the equation in terms of a common base, or we can use numerical methods to approximate the solution.

Q: What is the domain of the solution?

A: The domain of the solution is all real numbers x such that $-\ln(6) - 3\ln(2) \neq 0$. This means that the solution is valid for all real numbers x except for those that make the denominator of the expression equal to zero.

Q: Can we use the solution to solve other equations involving exponential terms?

A: Yes, we can use the solution to solve other equations involving exponential terms. The technique of taking the logarithm of both sides and using the properties of logarithms can be applied to a wide range of equations.

Conclusion

In this article, we have answered some common questions about solving the equation $6^{1-x} = 2^{3x+1}$ for x. We have discussed the significance of taking the logarithm of both sides, the relationship between the logarithm and the exponential function, and the domain of the solution. We have also shown that the solution can be used to solve other equations involving exponential terms.

Final Answer

The final answer is:

$$x = \frac{\ln\left(\frac{1}{3}\right)}{\ln(3) + 3\ln(2)}$$

This is the solution to the equation $6^{1-x} = 2^{3x+1}$ for x.