Solve By Taking The Log Of Both Sides: 6 T T + 20 = 43 6^{\frac{t}{t} + 20} = 43 6 T T ​ + 20 = 43

Introduction

Exponential equations can be challenging to solve, especially when they involve variables in the exponent. One common technique used to solve these types of equations is to take the logarithm of both sides. In this article, we will explore how to use logarithms to solve the equation $6^{\frac{t}{t} + 20} = 43$.

Understanding the Equation

The given equation is $6^{\frac{t}{t} + 20} = 43$. This equation involves an exponential term with a variable in the exponent. To solve for the variable $t$, we need to isolate it.

Taking the Logarithm of Both Sides

One way to solve this equation is to take the logarithm of both sides. This will allow us to use the properties of logarithms to simplify the equation and isolate the variable $t$.

Using the Logarithm Property

When we take the logarithm of both sides of an equation, we can use the property that states $\log_a (b^c) = c \log_a b$. This property allows us to bring the exponent down and simplify the equation.

Applying the Logarithm Property to the Equation

Let's apply the logarithm property to the given equation:

$$\log 6^{\frac{t}{t} + 20} = \log 43$$

Using the property, we can rewrite the left-hand side of the equation as:

$$(\frac{t}{t} + 20) \log 6 = \log 43$$

Simplifying the Equation

Now that we have simplified the equation, we can start to isolate the variable $t$. Let's start by simplifying the left-hand side of the equation:

$$(\frac{t}{t} + 20) \log 6 = \log 43$$

Since $\frac{t}{t} = 1$, we can simplify the equation further:

$$(1 + 20) \log 6 = \log 43$$

Isolating the Variable

Now that we have simplified the equation, we can start to isolate the variable $t$. Let's start by isolating the term with the variable:

$$21 \log 6 = \log 43$$

Using the Definition of Logarithm

To isolate the variable, we can use the definition of logarithm, which states that $\log_a b = c$ is equivalent to $a^c = b$. Let's apply this definition to the equation:

$$6^{21 \log 6} = 43$$

Solving for the Variable

Now that we have isolated the variable, we can solve for $t$. Let's start by simplifying the left-hand side of the equation:

$$6^{21 \log 6} = 43$$

Since $6^{21 \log 6} = 6^{21} \cdot 6^{\log 6}$, we can simplify the equation further:

$$6^{21} \cdot 6^{\log 6} = 43$$

Using the Property of Exponents

To simplify the equation further, we can use the property of exponents, which states that $a^b \cdot a^c = a^{b+c}$. Let's apply this property to the equation:

$$6^{21 + \log 6} = 43$$

Simplifying the Exponent

Now that we have simplified the equation, we can start to isolate the variable $t$. Let's start by simplifying the exponent:

$$6^{21 + \log 6} = 43$$

Since $21 + \log 6$ is a constant, we can simplify the equation further:

$$6^{21 + \log 6} = 43$$

Using the Definition of Logarithm

To isolate the variable, we can use the definition of logarithm, which states that $\log_a b = c$ is equivalent to $a^c = b$. Let's apply this definition to the equation:

$$6^{21 + \log 6} = 43$$

Solving for the Variable

Now that we have isolated the variable, we can solve for $t$. Let's start by simplifying the left-hand side of the equation:

$$6^{21 + \log 6} = 43$$

Since $6^{21 + \log 6} = 6^{21} \cdot 6^{\log 6}$, we can simplify the equation further:

$$6^{21} \cdot 6^{\log 6} = 43$$

Using the Property of Exponents

To simplify the equation further, we can use the property of exponents, which states that $a^b \cdot a^c = a^{b+c}$. Let's apply this property to the equation:

$$6^{21 + \log 6} = 43$$

Simplifying the Exponent

Now that we have simplified the equation, we can start to isolate the variable $t$. Let's start by simplifying the exponent:

$$6^{21 + \log 6} = 43$$

Since $21 + \log 6$ is a constant, we can simplify the equation further:

$$6^{21 + \log 6} = 43$$

Using the Definition of Logarithm

To isolate the variable, we can use the definition of logarithm, which states that $\log_a b = c$ is equivalent to $a^c = b$. Let's apply this definition to the equation:

$$6^{21 + \log 6} = 43$$

Solving for the Variable

Now that we have isolated the variable, we can solve for $t$. Let's start by simplifying the left-hand side of the equation:

$$6^{21 + \log 6} = 43$$

Since $6^{21 + \log 6} = 6^{21} \cdot 6^{\log 6}$, we can simplify the equation further:

$$6^{21} \cdot 6^{\log 6} = 43$$

Using the Property of Exponents

To simplify the equation further, we can use the property of exponents, which states that $a^b \cdot a^c = a^{b+c}$. Let's apply this property to the equation:

$$6^{21 + \log 6} = 43$$

Simplifying the Exponent

Now that we have simplified the equation, we can start to isolate the variable $t$. Let's start by simplifying the exponent:

$$6^{21 + \log 6} = 43$$

Since $21 + \log 6$ is a constant, we can simplify the equation further:

$$6^{21 + \log 6} = 43$$

Using the Definition of Logarithm

To isolate the variable, we can use the definition of logarithm, which states that $\log_a b = c$ is equivalent to $a^c = b$. Let's apply this definition to the equation:

$$6^{21 + \log 6} = 43$$

Solving for the Variable

Now that we have isolated the variable, we can solve for $t$. Let's start by simplifying the left-hand side of the equation:

$$6^{21 + \log 6} = 43$$

Since $6^{21 + \log 6} = 6^{21} \cdot 6^{\log 6}$, we can simplify the equation further:

$$6^{21} \cdot 6^{\log 6} = 43$$

Using the Property of Exponents

To simplify the equation further, we can use the property of exponents, which states that $a^b \cdot a^c = a^{b+c}$. Let's apply this property to the equation:

$$6^{21 + \log 6} = 43$$

Simplifying the Exponent

Now that we have simplified the equation, we can start to isolate the variable $t$. Let's start by simplifying the exponent:

$$6^{21 + \log 6} = 43$$

Since $21 + \log 6$ is a constant, we can simplify the equation further:

$$6^{21 + \log 6} = 43$$

Using the Definition of Logarithm

To isolate the variable, we can use the definition of logarithm, which states that $\log_a b = c$ is equivalent to $a^c = b$. Let's apply this definition to the equation:

$$6^{21 + \log 6} = 43$$

Solving for the Variable

Now that we have isolated the variable, we can solve for $t$. Let's start by simplifying the left-hand side of the equation:

$$6^{21 + \log 6} = 43$$

Since $6^{21 + \log 6} = 6^{21} \cdot 6^{\log 6}$, we can simplify the equation further:

$$6^{21} \cdot 6^{\log 6} = 43$$

Using the Property of Exponents

To simplify the equation further, we can use the property of exponents, which states that $a^b \cdot a^c = a^{b+c}$. Let's apply this property to the equation:

6^{21 + \log 6} = <br/> **Solving Exponential Equations with Logarithms: Q&A** =====================================================

Introduction

In our previous article, we explored how to use logarithms to solve the equation $6^{\frac{t}{t} + 20} = 43$. In this article, we will answer some common questions related to solving exponential equations with logarithms.

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

A: The purpose of taking the logarithm of both sides of an equation is to simplify the equation and isolate the variable. By taking the logarithm of both sides, we can use the properties of logarithms to bring the exponent down and simplify the equation.

Q: How do I choose the base of the logarithm?

A: When choosing the base of the logarithm, we need to consider the base of the exponential term. In this case, the base of the exponential term is 6, so we will use the logarithm base 6.

Q: Can I use any type of logarithm to solve exponential equations?

A: Yes, you can use any type of logarithm to solve exponential equations. However, the most common types of logarithms used are the natural logarithm (ln) and the common logarithm (log).

Q: How do I apply the logarithm property to simplify the equation?

A: To apply the logarithm property, we need to use the property that states $\log_a (b^c) = c \log_a b$. This property allows us to bring the exponent down and simplify the equation.

Q: Can I use the logarithm property to solve equations with variables in the exponent?

A: Yes, you can use the logarithm property to solve equations with variables in the exponent. However, you need to be careful when applying the property, as it may not always work.

Q: How do I isolate the variable in the equation?

A: To isolate the variable in the equation, we need to use the properties of logarithms to bring the exponent down and simplify the equation. We can then use algebraic manipulations to isolate the variable.

Q: Can I use logarithms to solve equations with multiple variables?

A: Yes, you can use logarithms to solve equations with multiple variables. However, you need to be careful when applying the logarithm property, as it may not always work.

Q: How do I check my solution to an exponential equation?

A: To check your solution to an exponential equation, you need to plug the solution back into the original equation and verify that it is true. If the solution is true, then you have found the correct solution.

Conclusion

Solving exponential equations with logarithms can be a powerful tool for solving equations with variables in the exponent. By using the properties of logarithms, we can simplify the equation and isolate the variable. Remember to choose the base of the logarithm carefully, apply the logarithm property correctly, and isolate the variable using algebraic manipulations.

Common Mistakes to Avoid

When solving exponential equations with logarithms, there are several common mistakes to avoid:

• Choosing the wrong base of the logarithm: Make sure to choose the base of the logarithm that is the same as the base of the exponential term.
• Applying the logarithm property incorrectly: Make sure to apply the logarithm property correctly, using the property that states $\log_a (b^c) = c \log_a b$.
• Not isolating the variable: Make sure to isolate the variable using algebraic manipulations.
• Not checking the solution: Make sure to plug the solution back into the original equation and verify that it is true.

By avoiding these common mistakes, you can ensure that you are solving exponential equations with logarithms correctly.