## Algebra 2 logarithmic

**Introduction :**

Algebra is one of the most basic element of mathematics in which, we switch from basic arithmetic to variables. Algebra covers a large number of subdivisions like polynomials, rational, exponents, logarithms, expressions etc under it. Algebra 2 is a sub division in which, certain topics are covered from algebra. Logarithmic problems comes under algebra 2. Logarithms function is a very common function and it plays a vital role in problems involving variable quantities and exponents. Logarithm of a certain number represents the value which is multiplied by 10 for the number of times given into the logarithmic function. For example: Consider log 6, it represents 1000000, similarly log 1000 represents 3.

Basic laws in Logarithm:

Solving the algebra 2 logarithmic type problem requires certain laws to to be followed to make it easier and reduce the given form to simple form. Those basic rules to be followed are given away as follows,

**Product Rule:**

**Quotient Rule**

**Power Rule**

**log a x n = n log a x**

**Change of Base Rule**

- I
**f a = b y,**then y = log b (a) - When the base and the value are same
**(log 10 10)**, it is equal to 1. **Log 0**to any base is equal to**1**.**Log 1**to any base is equal to**0**.

**Note:**The commonly used base is 10, and if no base is mentioned, then it is considered as log with base 10.Here some example problems for logarithmic function using the above rules.

Examples on algebra 2 logarithmic problems:

**Example 1:**

**Solve logarithmic function 2 log3 6 + log3 5 – log3 180**

**Solution:**

2 log3 6 + log3 5 – log3 180

(Using the power rule 2 comes as the power of 6),

= log3 62 + log3 5 – log3 180

= log3 36 + log3 5 – log3 180

(Using the product rule for the first two terms)

= log3 (36*5) - log3 180

(Using the quotient rule)

= log3 ((36*5)/180)

= log3 1

= 0

**Example 2:**

**Evaluate log5 25 + log3 243 – 2log4 2**

**Solution:**

log5 25 + log3 243 – 2 log4 2

= log5 52 + log3 35 – log4 22

(Using the power rule for all the terms)

= 2log5 5 + 5log3 3 – log4 4

= 2(1) + 5(1) – (1)

= 2+5-1

= 7-1 =6

**Example 3:**Solving logarithmic function 2 log3 27 + log3 81 – 3 log3 9

**Solution:**

= 2 log3 33 + log3 34 – 3 log3 32

= 3*2log3 3 + 4log3 3 – 2*3log3 3

(Using the power rule to all the terms),

= 6log3 3 +4log3 3– 6log3 3

=6(1) + 4(1) -6(1)

=10-6

=4.

**Algebra 2 Practice problems:**

1. Solving logarithmic function for 2 log3 81 + log3 243 – 3 log3 27

2. Evaluate log5 125 + log3 729 + 2log4 64

**Answer:**

1. 8

2. 5

I like to share this Algebra 2 Answers with you all through my blog.