Question

Prove the following:

`2 log  15/4 + log  81/5 - 3 log  9/4 + log 100 = 3 + log 2`

Answer 1

Given: `2 log  (15/4) + log  (81/5) - 3 log  (9/4) + log 100`

To Prove: 3 + log 2

Proof (Step-wise):

1. Apply power and product/quotient laws of logarithms:

`2 log (15/4) + log(81/5) - 3 log(9/4) + log 100` 

= `log (15/4)^2 + log(81/5) + log (9/4)^(-3) + log 100` 

= `log (15/4)^2 xx (81/5) xx (9/4)^(-3) xx 100`

2. Combine and rewrite as a single fraction:

`(15/4)^2 xx (81/5) xx (9/4)^(−3) xx 100`

= `(225/16) xx (81/5) xx (1/(9/4))^3 xx 100` 

= `(225/16) xx (81/5) xx (64/729) xx 100`

3. Factor into primes and cancel common powers:

225 = 3² × 5²

81 = 3⁴

64 = 2⁶

729 = 3⁶

100 = 2² × 5²

16 = 2⁴

Multiply numerators:

`2^(6 + 2) = 2^8` 

`3^(2 + 4) = 3^6` 

`5^(2 + 2) = 5^4` 

Denominator:

2⁴ × 3⁶ × 5¹ 

Cancel 3⁶; remaining: `2^(8 - 4) = 2^4` and `5^(4 - 1) = 5^3`. 

So the product

= 2⁴ × 5³

= 16 × 125

= 2000

4. Take log: original expression

= log 2000 

= log (2 × 1000) 

= log 2 + log 1000

= log 2 + 3

`2 log (15/4) + log (81/5) − 3 log (9/4) + log 100 = 3 + log 2`