The logarithm of 1 with base a is 0

Oct 13, 2025 by magicalceylontours.com 50 views

Introduction

Solving for x in an exponential equation can be a challenging task, especially when dealing with logarithmic functions. In this article, we will explore the correct form to solve for x in a given equation and provide step-by-step solutions to each option.

Understanding Exponential Equations

Exponential equations are equations that involve an exponential function, which is a function of the form $f(x) = a^x$ , where a is a positive real number. Exponential equations can be written in the form $a^x = b$ , where a and b are positive real numbers.

Option A: $15^{3x-5} = 87$

To solve for x in this equation, we can use the property of logarithms that states $\log_a{b} = c \iff a^c = b$ . We can take the logarithm of both sides of the equation with base 15 to get:

\log_{15}{15^{3x-5}} = \log_{15}{87}

Using the property of logarithms that states $\log_a{a^c} = c$ , we can simplify the left-hand side of the equation to get:

3x-5 = \log_{15}{87}

Now, we can add 5 to both sides of the equation to get:

3x = \log_{15}{87} + 5

Finally, we can divide both sides of the equation by 3 to get:

x = \frac{\log_{15}{87} + 5}{3}

This is the correct form to solve for x in the equation $15^{3x-5} = 87$ .

Option B: $\frac{\log_{15}(87) + 5}{3} = x$

This option is the same as the solution we obtained in Option A. Therefore, it is also a correct form to solve for x in the equation $15^{3x-5} = 87$ .

Option C: $\frac{\log_{15}(92)}{3} = x$

This option is incorrect because it does not solve the original equation $15^{3x-5} = 87$ . The logarithm of 92 with base 15 is not equal to the logarithm of 87 with base 15.

Option D: $\log_{87}(3x-5) = 15$

This option is incorrect because it does not solve the original equation $15^{3x-5} = 87$ . The logarithm of 3x-5 with base 87 is not equal to 15.

Option E: No solution

This option is incorrect because there is a solution to the equation $15^{3x-5} = 87$ . We obtained the solution in Option A.

Conclusion

In conclusion, the correct form to solve for x in the equation $15^{3x-5} = 87$ is:

x = \frac{\log_{15}{87} + 5}{3}

This solution can be obtained by taking the logarithm of both sides of the equation with base 15 and then simplifying the resulting equation.

Understanding Logarithmic Functions

Logarithmic functions are functions of the form $f(x) = \log_a{x}$ , where a is a positive real number. Logarithmic functions have several important properties, including the property that states $\log_a{b} = c \iff a^c = b$ .

Properties of Logarithmic Functions

Logarithmic functions have several important properties, including:

The logarithm of 1 with base a is 0: $\log_a{1} = 0$
The logarithm of a with base a is 1: $\log_a{a} = 1$
The logarithm of a product is the sum of the logarithms: $\log_a{bc} = \log_a{b} + \log_a{c}$
The logarithm of a quotient is the difference of the logarithms: $\log_a{\frac{b}{c}} = \log_a{b} - \log_a{c}$

Using Logarithmic Functions to Solve Exponential Equations

Logarithmic functions can be used to solve exponential equations by taking the logarithm of both sides of the equation with a base that is equal to the base of the exponential function. This can help to simplify the equation and solve for the variable.

Example: Solving the Equation $2^x = 8$

To solve the equation $2^x = 8$ , we can take the logarithm of both sides of the equation with base 2 to get:

\log_2{2^x} = \log_2{8}

Using the property of logarithms that states $\log_a{a^c} = c$ , we can simplify the left-hand side of the equation to get:

x = \log_2{8}

Now, we can use the property of logarithms that states $\log_a{b} = c \iff a^c = b$ to solve for x:

2^x = 8 \iff x = \log_2{8}

Therefore, the solution to the equation $2^x = 8$ is $x = \log_2{8}$ .

Conclusion

In conclusion, logarithmic functions can be used to solve exponential equations by taking the logarithm of both sides of the equation with a base that is equal to the base of the exponential function. This can help to simplify the equation and solve for the variable.

Final Answer

The final answer to the problem is:

x = \frac{\log_{15}{87} + 5}{3}$$<br/> # Q&A: Solving Exponential Equations with Logarithmic Functions

Introduction

In our previous article, we discussed how to solve exponential equations using logarithmic functions. We also provided a step-by-step solution to the equation $15^{3x-5} = 87$ . In this article, we will answer some frequently asked questions about solving exponential equations with logarithmic functions.

Q: What is the difference between an exponential equation and a logarithmic equation?

A: An exponential equation is an equation that involves an exponential function, which is a function of the form $f(x) = a^x$ , where a is a positive real number. A logarithmic equation, on the other hand, is an equation that involves a logarithmic function, which is a function of the form $f(x) = \log_a{x}$ , where a is a positive real number.

Q: How do I know which base to use when solving an exponential equation with logarithmic functions?

A: When solving an exponential equation with logarithmic functions, you should use the base that is equal to the base of the exponential function. For example, if the equation is $2^x = 8$ , you should use the base 2 when taking the logarithm of both sides of the equation.

Q: Can I use any base when solving an exponential equation with logarithmic functions?

A: No, you should only use the base that is equal to the base of the exponential function. Using a different base can lead to incorrect solutions.

Q: How do I simplify the equation after taking the logarithm of both sides?

A: After taking the logarithm of both sides of the equation, you can simplify the equation by using the properties of logarithms. For example, if the equation is $\log_2{2^x} = \log_2{8}$ , you can simplify the equation to $x = \log_2{8}$ .

Q: Can I use logarithmic functions to solve equations with negative exponents?

A: Yes, you can use logarithmic functions to solve equations with negative exponents. However, you should be careful when simplifying the equation, as the negative exponent can affect the solution.

Q: How do I know if the solution to an exponential equation is valid?

A: To determine if the solution to an exponential equation is valid, you should check if the solution satisfies the original equation. If the solution does not satisfy the original equation, then it is not a valid solution.

Q: Can I use logarithmic functions to solve equations with fractional exponents?

A: Yes, you can use logarithmic functions to solve equations with fractional exponents. However, you should be careful when simplifying the equation, as the fractional exponent can affect the solution.

Q: How do I choose the correct logarithmic function to use when solving an exponential equation?

A: When choosing the correct logarithmic function to use when solving an exponential equation, you should consider the base of the exponential function and the properties of logarithms. You should also consider the properties of the equation, such as the presence of negative exponents or fractional exponents.

Q: Can I use logarithmic functions to solve equations with complex numbers?

A: Yes, you can use logarithmic functions to solve equations with complex numbers. However, you should be careful when simplifying the equation, as the complex numbers can affect the solution.

Q: How do I know if the solution to an exponential equation with complex numbers is valid?

A: To determine if the solution to an exponential equation with complex numbers is valid, you should check if the solution satisfies the original equation. If the solution does not satisfy the original equation, then it is not a valid solution.

Conclusion

In conclusion, logarithmic functions can be used to solve exponential equations by taking the logarithm of both sides of the equation with a base that is equal to the base of the exponential function. By understanding the properties of logarithmic functions and how to simplify the equation, you can solve exponential equations with ease.

Final Answer

The final answer to the problem is:

x = \frac{\log_{15}{87} + 5}{3} </span></p>