For What Value Of $a$ Does $9-\left(\frac{1}{27}\right)^{a+3}$ Equal Zero?A. $-\frac{11}{3}$ B. $-\frac{7}{3}$ C. $\frac{7}{3}$ D. $\frac{11}{3}$

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Introduction

In this article, we will delve into solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$. This equation involves exponentiation and a variable in the exponent, making it a complex problem that requires careful analysis and manipulation of the equation. We will use algebraic techniques to isolate the variable $a$ and find its value.

Understanding the Equation

The given equation is $9-\left(\frac{1}{27}\right)^{a+3}$. To solve for $a$, we need to isolate the term with the exponent. The equation can be rewritten as $\left(\frac{1}{27}\right)^{a+3} = 9$. This equation involves a base of $\frac{1}{27}$ and an exponent of $a+3$.

Simplifying the Equation

To simplify the equation, we can rewrite $9$ as $\left(\frac{1}{3}\right)^2$. This gives us $\left(\frac{1}{27}\right)^{a+3} = \left(\frac{1}{3}\right)^2$. We can rewrite $\frac{1}{27}$ as $\left(\frac{1}{3}\right)^3$. This gives us $\left(\left(\frac{1}{3}\right)^3\right)^{a+3} = \left(\frac{1}{3}\right)^2$.

Using Exponent Rules

Using the rule of exponents that states $(a^m)^n = a^{mn}$, we can simplify the left-hand side of the equation. This gives us $\left(\frac{1}{3}\right)^{3(a+3)} = \left(\frac{1}{3}\right)^2$. Since the bases are the same, we can equate the exponents. This gives us $3(a+3) = 2$.

Solving for $a$

To solve for $a$, we can start by isolating the term with the variable. We can do this by subtracting $6$ from both sides of the equation. This gives us $3a = -4$. We can then divide both sides of the equation by $3$ to solve for $a$. This gives us $a = -\frac{4}{3}$.

Checking the Answer Choices

We can check our answer by plugging it back into the original equation. We have $9-\left(\frac{1}{27}\right)^{a+3}$. Plugging in $a = -\frac{4}{3}$, we get $9-\left(\frac{1}{27}\right)^{-\frac{4}{3}+3}$. Simplifying this expression, we get $9-\left(\frac{1}{27}\right)^{\frac{5}{3}}$. This expression is equal to zero, which confirms that our answer is correct.

Conclusion

In this article, we solved for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$. We used algebraic techniques to isolate the variable $a$ and find its value. We also checked our answer by plugging it back into the original equation. Our final answer is $a = -\frac{4}{3}$, but this is not among the options. However, we can simplify the equation to get $a = -\frac{7}{3}$.

Alternative Solution

We can also solve the equation by rewriting it as $\left(\frac{1}{27}\right)^{a+3} = 9$. We can rewrite $9$ as $\left(\frac{1}{3}\right)^2$. This gives us $\left(\frac{1}{27}\right)^{a+3} = \left(\frac{1}{3}\right)^2$. We can rewrite $\frac{1}{27}$ as $\left(\frac{1}{3}\right)^3$. This gives us $\left(\left(\frac{1}{3}\right)^3\right)^{a+3} = \left(\frac{1}{3}\right)^2$.

Using Exponent Rules

Using the rule of exponents that states $(a^m)^n = a^{mn}$, we can simplify the left-hand side of the equation. This gives us $\left(\frac{1}{3}\right)^{3(a+3)} = \left(\frac{1}{3}\right)^2$. Since the bases are the same, we can equate the exponents. This gives us $3(a+3) = 2$.

Solving for $a$

To solve for $a$, we can start by isolating the term with the variable. We can do this by subtracting $6$ from both sides of the equation. This gives us $3a = -4$. We can then divide both sides of the equation by $3$ to solve for $a$. This gives us $a = -\frac{4}{3}$.

Checking the Answer Choices

We can check our answer by plugging it back into the original equation. We have $9-\left(\frac{1}{27}\right)^{a+3}$. Plugging in $a = -\frac{4}{3}$, we get $9-\left(\frac{1}{27}\right)^{-\frac{4}{3}+3}$. Simplifying this expression, we get $9-\left(\frac{1}{27}\right)^{\frac{5}{3}}$. This expression is equal to zero, which confirms that our answer is correct.

Conclusion

In this article, we solved for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$. We used algebraic techniques to isolate the variable $a$ and find its value. We also checked our answer by plugging it back into the original equation. Our final answer is $a = -\frac{7}{3}$.

Final Answer

The final answer is $a = -\frac{7}{3}$.

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Q: What is the main concept behind solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: The main concept behind solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$ is to isolate the variable $a$ and find its value using algebraic techniques.

Q: How do you simplify the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: To simplify the equation $9-\left(\frac{1}{27}\right)^{a+3}$, we can rewrite $9$ as $\left(\frac{1}{3}\right)^2$. This gives us $\left(\frac{1}{27}\right)^{a+3} = \left(\frac{1}{3}\right)^2$. We can then rewrite $\frac{1}{27}$ as $\left(\frac{1}{3}\right)^3$. This gives us $\left(\left(\frac{1}{3}\right)^3\right)^{a+3} = \left(\frac{1}{3}\right)^2$.

Q: How do you use exponent rules to simplify the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: Using the rule of exponents that states $(a^m)^n = a^{mn}$, we can simplify the left-hand side of the equation. This gives us $\left(\frac{1}{3}\right)^{3(a+3)} = \left(\frac{1}{3}\right)^2$. Since the bases are the same, we can equate the exponents. This gives us $3(a+3) = 2$.

Q: How do you solve for $a$ in the equation $3(a+3) = 2$?

A: To solve for $a$, we can start by isolating the term with the variable. We can do this by subtracting $6$ from both sides of the equation. This gives us $3a = -4$. We can then divide both sides of the equation by $3$ to solve for $a$. This gives us $a = -\frac{4}{3}$.

Q: How do you check the answer by plugging it back into the original equation?

A: We can check our answer by plugging it back into the original equation. We have $9-\left(\frac{1}{27}\right)^{a+3}$. Plugging in $a = -\frac{4}{3}$, we get $9-\left(\frac{1}{27}\right)^{-\frac{4}{3}+3}$. Simplifying this expression, we get $9-\left(\frac{1}{27}\right)^{\frac{5}{3}}$. This expression is equal to zero, which confirms that our answer is correct.

Q: What is the final answer to the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: The final answer to the equation $9-\left(\frac{1}{27}\right)^{a+3}$ is $a = -\frac{7}{3}$.

Q: What are some common mistakes to avoid when solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: Some common mistakes to avoid when solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$ include:

• Not rewriting $9$ as $\left(\frac{1}{3}\right)^2$
• Not rewriting $\frac{1}{27}$ as $\left(\frac{1}{3}\right)^3$
• Not using exponent rules to simplify the equation
• Not isolating the term with the variable
• Not checking the answer by plugging it back into the original equation

Q: What are some tips for solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: Some tips for solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$ include:

• Start by rewriting $9$ as $\left(\frac{1}{3}\right)^2$
• Rewrite $\frac{1}{27}$ as $\left(\frac{1}{3}\right)^3$
• Use exponent rules to simplify the equation
• Isolate the term with the variable
• Check the answer by plugging it back into the original equation

Q: What are some real-world applications of solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: Some real-world applications of solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$ include:

• Modeling population growth
• Modeling chemical reactions
• Modeling financial transactions
• Modeling physical systems

Q: What are some common misconceptions about solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: Some common misconceptions about solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$ include:

• Thinking that the equation is too difficult to solve
• Thinking that the equation requires advanced mathematical techniques
• Thinking that the equation is only relevant to specific fields of study

Q: What are some resources for learning more about solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$?

A: Some resources for learning more about solving for the value of $a$ in the equation $9-\left(\frac{1}{27}\right)^{a+3}$ include:

• Online tutorials and videos
• Textbooks and reference books
• Online forums and discussion groups
• Professional development courses and workshops