Exercise 1 Solve For The Unknown In Each Case:(i) { \frac{\log X}{\log 2}=2$}$(ii) { \log 4+\log X=2$}$(iii) ${ 3 \ln X=6\$} (iv) { \log (0.01)^x=3$}$(v) { \ln X-\ln (x-2)=4 \ln \sqrt{e}$}$(vi)

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Introduction

Logarithmic equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will delve into the world of logarithmic equations and provide a step-by-step guide on how to solve them. We will cover six different cases, each with its unique challenges and solutions.

Case (i): Solving for x in {\frac{\log x}{\log 2}=2$}$

To solve this equation, we can start by using the properties of logarithms. Specifically, we can use the fact that {\log a^b=b\log a$}$. Applying this property to the given equation, we get:

{\frac{\log x}{\log 2}=2$}$

{\log x=2\log 2$}$

{\log x=\log 2^2$}$

{\log x=\log 4$}$

Since the logarithm function is one-to-one, we can equate the arguments of the logarithms:

{x=4$}$

Therefore, the solution to the equation is {x=4$}$.

Case (ii): Solving for x in {\log 4+\log x=2$}$

To solve this equation, we can use the property of logarithms that states {\log a+\log b=\log ab$}$. Applying this property to the given equation, we get:

{\log 4+\log x=2$}$

{\log (4x)=2$}$

${4x=10^2\$}

${4x=100\$}

{x=\frac{100}{4}$}$

{x=25$}$

Therefore, the solution to the equation is {x=25$}$.

Case (iii): Solving for x in ${3 \ln x=6\$}

To solve this equation, we can start by isolating the logarithmic term:

${3 \ln x=6\$}

{\ln x=\frac{6}{3}$}$

{\ln x=2$}$

Since the natural logarithm function is one-to-one, we can equate the arguments of the logarithms:

{x=e^2$}$

Therefore, the solution to the equation is {x=e^2$}$.

Case (iv): Solving for x in {\log (0.01)^x=3$}$

To solve this equation, we can use the property of logarithms that states {\log a^b=b\log a$}$. Applying this property to the given equation, we get:

{\log (0.01)^x=3$}$

{x\log 0.01=3$}$

{x\log \frac{1}{100}=3$}$

{x(-2\log 10)=3$}$

{x(-2)=3$}$

{x=-\frac{3}{2}$}$

Therefore, the solution to the equation is {x=-\frac{3}{2}$}$.

Case (v): Solving for x in {\ln x-\ln (x-2)=4 \ln \sqrt{e}$}$

To solve this equation, we can start by using the properties of logarithms. Specifically, we can use the fact that {\ln a-\ln b=\ln \frac{a}{b}$}$. Applying this property to the given equation, we get:

{\ln x-\ln (x-2)=4 \ln \sqrt{e}$}$

{\ln \frac{x}{x-2}=4 \ln \sqrt{e}$}$

{\ln \frac{x}{x-2}=2 \ln e$}$

{\ln \frac{x}{x-2}=\ln e^2$}$

Since the logarithm function is one-to-one, we can equate the arguments of the logarithms:

{\frac{x}{x-2}=e^2$}$

{x=e^2(x-2)$}$

{x=e2x-2e2$}$

{x(1-e2)=-2e2$}$

{x=\frac{-2e2}{1-e2}$}$

Therefore, the solution to the equation is {x=\frac{-2e2}{1-e2}$}$.

Conclusion

Q: What is a logarithmic equation?

A: A logarithmic equation is an equation that involves a logarithmic function. It is an equation that can be written in the form {\log_a x=b$}$, where {a$}$ is the base of the logarithm and {x$}$ is the argument of the logarithm.

Q: What are the common bases of logarithms?

A: The common bases of logarithms are ${10\$} (common logarithm) and {e$}$ (natural logarithm). The common logarithm is denoted by {\log$}$ and the natural logarithm is denoted by {\ln$}$.

Q: How do I solve a logarithmic equation?

A: To solve a logarithmic equation, you need to isolate the logarithmic term and then use the properties of logarithms to simplify the equation. Here are the steps:

  1. Isolate the logarithmic term.
  2. Use the properties of logarithms to simplify the equation.
  3. Equate the arguments of the logarithms.
  4. Solve for the variable.

Q: What are the properties of logarithms?

A: The properties of logarithms are:

  1. {\log a^b=b\log a$}$
  2. {\log a+\log b=\log ab$}$
  3. {\log a-\log b=\log \frac{a}{b}$}$
  4. {\log a^b=b\log a$}$

Q: How do I use the properties of logarithms to simplify an equation?

A: To use the properties of logarithms to simplify an equation, you need to identify the logarithmic terms and then apply the properties of logarithms to simplify the equation. Here are the steps:

  1. Identify the logarithmic terms.
  2. Apply the properties of logarithms to simplify the equation.
  3. Simplify the equation.

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

A: A logarithmic equation is an equation that involves a logarithmic function, while an exponential equation is an equation that involves an exponential function. For example, {\log x=2$}$ is a logarithmic equation, while {x^2=4$}$ is an exponential equation.

Q: How do I solve an exponential equation?

A: To solve an exponential equation, you need to isolate the exponential term and then use the properties of exponents to simplify the equation. Here are the steps:

  1. Isolate the exponential term.
  2. Use the properties of exponents to simplify the equation.
  3. Equate the arguments of the exponential function.
  4. Solve for the variable.

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

A: Some common mistakes to avoid when solving logarithmic equations are:

  1. Not isolating the logarithmic term.
  2. Not using the properties of logarithms to simplify the equation.
  3. Not equating the arguments of the logarithms.
  4. Not solving for the variable.

Conclusion

Logarithmic equations can be challenging to solve, but with practice and patience, you can become proficient in solving them. Remember to use the properties of logarithms to simplify the equations and isolate the variable. With this article, you should be able to answer frequently asked questions about logarithmic equations and solve them with ease.