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Learn logarithmic identities including product, quotient, power, and change of base formulas. Solve equations with practical real-world examples.

---

## Understanding Logarithmic Identities

Logarithmic identities are special properties that apply to all logarithmic functions. These properties are very helpful in simplifying calculations and solving complex logarithmic equations.

Before discussing logarithmic identities, let's recall that logarithms are the inverse of exponents. If $$b^x = a$$, then $$^b\log a = x$$.

Visible text: Before discussing logarithmic identities, let's recall that logarithms are the inverse of exponents. If , then .

## Basic Logarithmic Identities

### Product Identity

```math
^b\log(MN) = {^b\log M} + {^b\log N}
```

The logarithm of a product equals the sum of the logarithms of each number.

**Example:**

```math
^2\log(8 \times 4) = {^2\log 8} + {^2\log 4} = 3 + 2 = 5
```

### Quotient Identity

```math
^b\log\left(\frac{M}{N}\right) = {^b\log M} - {^b\log N}
```

The logarithm of a quotient equals the difference between the logarithm of the numerator and the logarithm of the denominator.

**Example:**

```math
^3\log\left(\frac{81}{9}\right) = {^3\log 81} - {^3\log 9} = 4 - 2 = 2
```

### Power Identity

```math
^b\log(M^p) = p \cdot {^b\log M}
```

The logarithm of a number raised to a power equals the power multiplied by the logarithm of that number.

**Example:**

```math
^2\log(4^3) = 3 \cdot {^2\log 4} = 3 \times 2 = 6
```

## Special Logarithmic Identities

### Change of Base

```math
^b\log M = \frac{^a\log M}{^a\log b}
```

This identity allows us to change the logarithm base as needed.

**Example:**

```math
^3\log 9 = \frac{^{10}\log 9}{^{10}\log 3} = \frac{0.954}{0.477} = 2
```

### Equality Identity

If $$^b\log M = {^b\log N}$$, then $$M = N$$

Visible text: If , then

Two numbers that have the same logarithmic value (with the same base) must be the same number.

### Inequality Identity

- If $$b > 1$$ and $$^b\log M < {^b\log N}$$, then $$M < N$$
- If $$0 < b < 1$$ and $$^b\log M < {^b\log N}$$, then $$M > N$$

Visible text: - If and , then 
- If and , then

## Applications of Logarithmic Identities

### Simplifying Expressions

Simplify $$^2\log 32 + {^2\log 8} - {^2\log 4}$$

Visible text: Simplify

**Solution:**

Component: MathContainer
Children:

```math
^2\log 32 + {^2\log 8} - {^2\log 4}
```

```math
= {^2\log(32 \times 8)} - {^2\log 4}
```

```math
= {^2\log 256} - {^2\log 4}
```

```math
= {^2\log\left(\frac{256}{4}\right)}
```

```math
= {^2\log 64} = 6
```

### Solving Equations

Find the value of $$x$$ if $$^3\log x + {^3\log 9} = {^3\log 81}$$

Visible text: Find the value of if

**Solution:**

Component: MathContainer
Children:

```math
^3\log x + {^3\log 9} = {^3\log 81}
```

```math
^3\log(x \times 9) = {^3\log 81}
```

```math
x \times 9 = 81
```

```math
x = \frac{81}{9} = 9
```

## Real-Life Applications

### Richter Scale

Earthquake strength is measured using the Richter scale which is based on logarithms:

```math
R = \log\left(\frac{I}{I_0}\right)
```

Where:

- $$R$$ = Richter scale value
- $$I$$ = earthquake intensity
- $$I_0$$ = reference intensity (zero level)

Visible text: - = Richter scale value
- = earthquake intensity
- = reference intensity (zero level)

**Example:** An earthquake that occurred in Haiti in $$2010$$ had an intensity of $$10^7\text{ times}$$ compared to zero-level earthquakes. What is the Richter scale strength of that earthquake?

Visible text: **Example:** An earthquake that occurred in Haiti in had an intensity of compared to zero-level earthquakes. What is the Richter scale strength of that earthquake?

**Solution:**

Component: MathContainer
Children:

```math
R = \log\left(\frac{I}{I_0}\right)
```

```math
= \log\left(\frac{10^7 I_0}{I_0}\right)
```

```math
= \log(10^7)
```

```math
= 7
```

Therefore, the earthquake in Haiti in $$2010$$ had a strength of $$7$$ on the Richter scale.

Visible text: Therefore, the earthquake in Haiti in had a strength of on the Richter scale.

### Battery Charging

Battery charging time can be calculated using the logarithmic formula:

```math
t = -\frac{1}{k}\ln\left(1 - \frac{C}{C_0}\right)
```

Where:

- $$t$$ = charging time (in minutes)
- $$k$$ = charging constant
- $$C$$ = desired capacity
- $$C_0$$ = maximum capacity

Visible text: - = charging time (in minutes)
- = charging constant
- = desired capacity
- = maximum capacity

**Example:** Determine the time required to charge a battery from empty to $$90\%$$ full. Assume $$k = 0.02$$.

Visible text: **Example:** Determine the time required to charge a battery from empty to full. Assume .

**Solution:**

Component: MathContainer
Children:

```math
t = -\frac{1}{k}\ln\left(1 - \frac{C}{C_0}\right)
```

```math
= -\frac{1}{0.02}\ln\left(1 - \frac{0.9C_0}{C_0}\right)
```

```math
= -50\ln(1 - 0.9)
```

```math
= -50\ln(0.1)
```

```math
\approx 115.13
```

Therefore, the charging time is approximately $$115 \text{ minutes}$$.

Visible text: Therefore, the charging time is approximately .

### Car Price Depreciation

Logarithmic functions are also used for modeling decay/depreciation with the formula:

```math
H(t) = ce^{kt}
```

where $$H(t)$$ is the value at time $$t$$.

Visible text: where is the value at time .

**Example:** At any given time, the price of a used car is not proportional to its current price. If a new car costs $$200 \text{ million rupiah}$$ and after $$5 \text{ years}$$ becomes $$100 \text{ million rupiah}$$, determine the car's price after $$10 \text{ years}$$ of use.

Visible text: **Example:** At any given time, the price of a used car is not proportional to its current price. If a new car costs and after becomes , determine the car's price after of use.

**Solution:**

Component: MathContainer
Children:

```math
H(0) = 200 \text{ million, so} 200 = ce^0 = c
```

```math
H(5) = 100 \text{ million, so} 100 = 200e^{5k}
```

```math
e^{5k} = \frac{1}{2} \text{, therefore } 5k = \ln\left(\frac{1}{2}\right)
```

```math
k = \frac{1}{5}\ln\left(\frac{1}{2}\right) = -0.1386
```

From these results, the car's price at any time $$t$$ is:

Visible text: From these results, the car's price at any time is:

```math
H(t) = 200e^{-0.1386t}
```

Therefore, the car's price after $$10 \text{ years}$$ of use is:

Visible text: Therefore, the car's price after of use is:

```math
H(10) = 200e^{-0.1386(10)} = 200e^{-1.386} \approx 50 \text{ million rupiah}
```

## Exercises

**Problem** $$1$$

Visible text: **Problem**

Simplify: $$^5\log 125 + {^5\log 25} - {^5\log 5}$$

Visible text: Simplify:

**Problem** $$2$$

Visible text: **Problem**

If $$^2\log x = 3$$ and $$^2\log y = 5$$, find the value of $$^2\log(xy)$$

Visible text: If and , find the value of

**Problem** $$3$$

Visible text: **Problem**

Find the value of $$x$$ if $$^4\log(x-1) = {^4\log 16} - {^4\log 2}$$

Visible text: Find the value of if

### Answer Key

**Answer** $$1$$

Visible text: **Answer**

Component: MathContainer
Children:

```math
^5\log 125 + {^5\log 25} - {^5\log 5}
```

```math
= 3 + 2 - 1 = 4
```

**Answer** $$2$$

Visible text: **Answer**

Component: MathContainer
Children:

```math
^2\log(xy) = {^2\log x} + {^2\log y}
```

```math
= 3 + 5 = 8
```

**Answer** $$3$$

Visible text: **Answer**

Component: MathContainer
Children:

```math
^4\log(x-1) = {^4\log 16} - {^4\log 2}
```

```math
^4\log(x-1) = {^4\log\left(\frac{16}{2}\right)}
```

```math
^4\log(x-1) = {^4\log 8}
```

```math
x - 1 = 8
```

```math
x = 9
```