## Problems on Logarithm Rules

#### Algebra > Logarithm > Problems on Log Rules

4.Use Log Rules to simplify the Following questions:

1. log a a3       2. log 3 3√(81)

Solution 1:

Let us use the third Rule of Logarithm given above:

log a (p)n = n log a p

log a a3 = 3 log a a

= 3 × 1 = 3 ( because log a a = 1)

Solution 2:

First of all,

3√(81) = 3√(34) = (34)1/3 = 34/3

So, log 3 3√(81) = log 3 34/3

= (4/3) × log 3 3 = (4/3) ×1 (because log 3 3 = 1)

= 4 / 3.

4. Find the values of the following

1. 33 – log 37       2. [ 8log 2 3√11]1/3

Solution1:

First recall laws of exponents:

am – n = am / an

So, 33 – log 37

= (3)3 / (3)log 37

= (27) / (3)log 37 ……..(1)

Now, apply the formula: alog a p = p in the denominator for 3 log 37

So, 3log 37 = 7

Put this value 7 in the denominator in (1), we get

33 – log 37 = 27/7

2. [8log 2 3√11]1/3

Solution2:

First of all,

8log 2 3√11

= (8) log 2 3√11         { because am is same as (a)m }

= (23)(log 2 3√11)

= (2)3log 2 3√11        {recall that (am)n = amn }

Now [8 log 2 3√11]1/3

= [(2)3log 2 3√11)]1/3

= [23log 2 3√11)]1/3

Recall laws of exponents again,

(am) 1/n = am/n

So, [23 log 2 3√11)] 1/3

= [23(log 2 3√11)/3]

= 2(log 2 3√11)       {because 3(log 2 3√11)/3 = log 2 3√11}

Now, apply apply the formula

a(log a p) = p on 2(log 2 3√11)

We get 2(log 2 3√11) = 3√11

Therefore,

[8log 2 3√11]1/3 = 3√11.