Hello, students of Math Lobby! This is a continuation of the ** previous article on Logarithm**, which we covered the definition of logarithm, its relation to exponentials, and the relationship between the laws of logarithm and indices.

So today in the second part, we will be covering the unique properties of logarithm and the different methods you can use to solve logarithm and exponential equations. Let’s begin!

__Unique properties of logarithm__

__Unique properties of logarithm__

There are certain unique properties of logarithm that we must know in order to help us solve logarithmic questions with greater ease. The first property that we know of, which we used in the proving portion of the laws of logarithm, is:

· **log***a*** a = 1**, which special cases includes **lg 10 = 1** and **ln e = 1**

Note: **lg** means log with a base of 10, and **ln** means log with a base of the mathematical constant, e, which is an irrational and transcendental number.

Reason: The reason why a logarithm that has the same base and argument is equals to 1 is cause if we were to look at it in the exponential form: **a = a^1**, which the equation will only hold true if the power of a is 1.

· **log***a*** 1 = 0**, which special cases includes **lg1 = 0 and ln1 = 0**

Reason: Let’s take a look at this in the exponential form: **1 = a^0**. This means that when the argument of a logarithm is 1, regardless of what the base is, the power of which **a** is being raised to must always be zero to keep the equation valid.

__Methods to solve logarithmic/exponential equations__

__Methods to solve logarithmic/exponential equations__

**For solving logarithmic equations:**

**Method 1:** Using the equality of logarithms to compare (when the constant term can be expressed in exponential form with the same base as the exponential term with the unknown)

**Example: Solve the equation log***3*** (x + 4) + log***3 ***(2 – x) = log***3*** (2 – 3x)**

Solution: log*3* (x + 4) + log*3* (2 – x) = log*3 *(2 – 3x)

log*3* (x + 4) (2 – x) = log*3* (2 – 3x)

**(x + 4) (2 – x) = 2 – 3x (Bases are the same (log3), hence arguments can be compared)**

8 – 2x – x^2 = 2 – 3x

x^2 – x – 6 = 0

(x + 2) (x – 3) = 0

x = -2 or 3

When x = 3, log3(2 – x) = log3(-1) is not defined, so therefore, x = -2

**Method 2:** Conversion of the logarithmic form into the exponential form (when the log terms have the same base and there is a constant term)

**Example: Solve the equation 2 lg x – lg (x + 20) = 1**

Solution: 2 lg(x) – lg (x + 20) = 1

lg(x^2) – lg (x + 20) = 1

lg[(x^2)/ (x + 20)] = 1

(x^2)/ (x + 20) = 101

x^2 = 10 (x + 20)

x^2 – 10x – 200 = 0

(x + 10) (x – 20) = 0

x = -10 or 20

When x = -10, 2 lg(x) = 2 lg (-10) is not defined, therefore x = 20

**Method 3:** Using the Change of Base Law first (when the log terms have different bases)

After converting the log term to the same base, then use either Method 1 or 2.

**Example: Solve the equation log***2*** x = 25 log***x***2**

Solution: log*2* (x) = 25 log*x* (2)

= 25/ [log*2* (x)]

[log*2* (x)]^2 = 25

log*2* (x) = +/- 5

x = 25 or 2-5

= 25 or 1/ 25

= 32 or 1/32

**For solving exponential equations:**

**Method 1:** Using the equality of indices to compare (when the constant term can be expressed in exponential form with the same base as the exponential term with the unknown)

**Example: Solve 5^(x+2) = 1/25**

5(x+2) = 1/25

5(x+2) = 1/ (5^2)

5(x+2) = 5^-2

Comparing indices,

x + 2 = -2

Therefore, x = -4

**Method 2:** Conversion of the exponential form into the logarithmic form (when the constant term cannot be expressed in exponential form with the same base as the exponential term with the unknown, and when the base is **e** or **1**)

**Example: Solve the equation e^(3x-1) = 8**

e^(3x-1) = 8

3x – 1 = ln 8

x = (1 + ln 8)/ 3

= 1.03 (3 s.f)

**Method 3:** Take log on both sides (when the constant term cannot be expressed in exponential form with the same base as the exponential term with the unknown, and when the base is not **e** or **10**)

**Example: Solve the equation 5^(y+2) = 7**

5^(y+2) = 7

lg(5y+2) = lg (7)

(y + 2) lg (5) = lg (7)

y + 2 = lg (7)/ lg (5)

y = [lg (7)/ lg (5)] – 2

= -0.791 (3 s.f)

And that’s all for today, students! This sums up the chapter on logarithm and ** Math Lobby **hopes that after this article, you have a clear understanding on the topic of logarithm and its unique properties, and is equipped with the necessary skills and methods to deal with questions involving logarithmic and exponential equations!

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