Given Log ⁡ 5 2 ≈ 0.4397 \log _5 2 \approx 0.4397 Lo G 5 ​ 2 ≈ 0.4397 And Log ⁡ 5 9 ≈ 1.5608 \log _5 9 \approx 1.5608 Lo G 5 ​ 9 ≈ 1.5608 , Evaluate Each Expression.12. $\log _5 75$13. $\log _5 30$14. Log ⁡ 5 5 72 \log _5 \frac{5}{72} Lo G 5 ​ 72 5 ​

Introduction

Logarithms are a fundamental concept in mathematics, used to represent the power to which a base number must be raised to obtain a given value. In this article, we will evaluate three logarithmic expressions using the given values of $\log _5 2 \approx 0.4397$ and $\log _5 9 \approx 1.5608$. We will use the properties of logarithms to simplify the expressions and find their values.

Evaluating $\log _5 75$

To evaluate $\log _5 75$, we can use the property of logarithms that states $\log _a (bc) = \log _a b + \log _a c$. We can rewrite $75$ as $5^2 \cdot 3$, so we have:

$$\log _5 75 = \log _5 (5^2 \cdot 3) = \log _5 5^2 + \log _5 3 = 2 + \log _5 3$$

We can use the given value of $\log _5 9 \approx 1.5608$ to find the value of $\log _5 3$. Since $9 = 3^2$, we have:

$$\log _5 9 = \log _5 3^2 = 2 \log _5 3$$

Solving for $\log _5 3$, we get:

$$\log _5 3 = \frac{\log _5 9}{2} = \frac{1.5608}{2} = 0.7804$$

Now we can substitute this value back into our original expression:

$$\log _5 75 = 2 + \log _5 3 = 2 + 0.7804 = 2.7804$$

Evaluating $\log _5 30$

To evaluate $\log _5 30$, we can use the property of logarithms that states $\log _a (bc) = \log _a b + \log _a c$. We can rewrite $30$ as $5 \cdot 6$, so we have:

$$\log _5 30 = \log _5 (5 \cdot 6) = \log _5 5 + \log _5 6 = 1 + \log _5 6$$

We can use the given value of $\log _5 9 \approx 1.5608$ to find the value of $\log _5 6$. Since $9 = 3^2$ and $6 = 2 \cdot 3$, we have:

$$\log _5 9 = \log _5 (2 \cdot 3)^2 = 2 \log _5 (2 \cdot 3) = 2 (\log _5 2 + \log _5 3)$$

Solving for $\log _5 2 + \log _5 3$, we get:

$$\log _5 2 + \log _5 3 = \frac{\log _5 9}{2} = \frac{1.5608}{2} = 0.7804$$

Now we can substitute this value back into our original expression:

$$\log _5 6 = \log _5 2 + \log _5 3 = 0.7804$$

Now we can substitute this value back into our original expression:

$$\log _5 30 = 1 + \log _5 6 = 1 + 0.7804 = 1.7804$$

Evaluating $\log _5 \frac{5}{72}$

To evaluate $\log _5 \frac{5}{72}$, we can use the property of logarithms that states $\log _a \frac{b}{c} = \log _a b - \log _a c$. We can rewrite $\frac{5}{72}$ as $\frac{5}{2^3 \cdot 3^2}$, so we have:

$$\log _5 \frac{5}{72} = \log _5 \frac{5}{2^3 \cdot 3^2} = \log _5 5 - \log _5 (2^3 \cdot 3^2)$$

We can use the given value of $\log _5 2 \approx 0.4397$ to find the value of $\log _5 (2^3 \cdot 3^2)$. Since $2^3 \cdot 3^2 = 8 \cdot 9 = 72$, we have:

$$\log _5 (2^3 \cdot 3^2) = \log _5 72 = 3 \log _5 2 + 2 \log _5 3$$

Solving for $3 \log _5 2 + 2 \log _5 3$, we get:

$$3 \log _5 2 + 2 \log _5 3 = 3 (0.4397) + 2 (0.7804) = 1.3191 + 1.5608 = 2.8799$$

Now we can substitute this value back into our original expression:

$$\log _5 \frac{5}{72} = \log _5 5 - \log _5 (2^3 \cdot 3^2) = 1 - 2.8799 = -1.8799$$

Conclusion

In this article, we evaluated three logarithmic expressions using the given values of $\log _5 2 \approx 0.4397$ and $\log _5 9 \approx 1.5608$. We used the properties of logarithms to simplify the expressions and find their values. The values of the expressions are:

• $\log _5 75 \approx 2.7804$
• $\log _5 30 \approx 1.7804$
• $\log _5 \frac{5}{72} \approx -1.8799$

Introduction

In our previous article, we evaluated three logarithmic expressions using the given values of $\log _5 2 \approx 0.4397$ and $\log _5 9 \approx 1.5608$. We used the properties of logarithms to simplify the expressions and find their values. In this article, we will answer some frequently asked questions related to logarithmic expressions and provide additional examples to help you understand the concept better.

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

A: A logarithmic expression is an expression that represents the power to which a base number must be raised to obtain a given value. For example, $\log _5 2$ represents the power to which 5 must be raised to obtain 2. On the other hand, an exponential expression is an expression that represents a value raised to a power. For example, $5^2$ represents 5 raised to the power of 2.

Q: How do I evaluate a logarithmic expression with a base other than 10?

A: To evaluate a logarithmic expression with a base other than 10, you can use the change of base formula. The change of base formula states that $\log _a b = \frac{\log _c b}{\log _c a}$, where $a$, $b$, and $c$ are positive real numbers and $c \neq 1$. For example, to evaluate $\log _5 2$, you can use the change of base formula with base 10: $\log _5 2 = \frac{\log _{10} 2}{\log _{10} 5}$.

Q: Can I use a calculator to evaluate a logarithmic expression?

A: Yes, you can use a calculator to evaluate a logarithmic expression. Most calculators have a logarithm button that allows you to enter the base and the value, and it will give you the result. For example, to evaluate $\log _5 2$, you can enter $\log 2$ and then press the 5 button to get the result.

Q: How do I simplify a logarithmic expression with multiple terms?

A: To simplify a logarithmic expression with multiple terms, you can use the properties of logarithms. The properties of logarithms state that:

• $\log _a (bc) = \log _a b + \log _a c$
• $\log _a (b/c) = \log _a b - \log _a c$
• $\log _a (b^c) = c \log _a b$

For example, to simplify $\log _5 (2 \cdot 3)$, you can use the first property: $\log _5 (2 \cdot 3) = \log _5 2 + \log _5 3$.

Q: Can I use logarithmic expressions to solve equations?

A: Yes, you can use logarithmic expressions to solve equations. Logarithmic expressions can be used to solve equations that involve exponential functions. For example, to solve the equation $5^x = 2$, you can take the logarithm of both sides: $\log _5 5^x = \log _5 2$. This simplifies to $x = \log _5 2$.

Q: What are some common applications of logarithmic expressions?

A: Logarithmic expressions have many common applications in science, engineering, and finance. Some examples include:

• Calculating the pH of a solution in chemistry
• Modeling population growth in biology
• Analyzing stock prices in finance
• Calculating the magnitude of an earthquake in seismology

Conclusion

In this article, we answered some frequently asked questions related to logarithmic expressions and provided additional examples to help you understand the concept better. We also discussed some common applications of logarithmic expressions and how they can be used to solve equations. We hope this article has been helpful in your understanding of logarithmic expressions.