Question

Prove the following:

`log  35/33 - log  135/99 + log  24/7 = 3 log 2 - log 3`

Answer 1

Given: `log (35/33) - log (135/99) + log (24/7)`

To Prove: `log (35/33) - log (135/99) + log (24/7) = 3 log 2 - log 3`

Proof (Step-wise):

1. Use log subtraction as division:

`log (35/33) - log (135/99)`

= `log [(35/33) ÷ (135/99)]` 

= `log [(35/33) xx (99/135)]`

2. Simplify the fraction:

`(35/33) xx (99/135)`

= `(5 xx 7)/(3 xx 11) xx (9 xx 11)/(3^3 xx 5)` 

= `(5 xx 7 xx 9 xx 11)/(3 xx 11 xx 27 xx 5)` 

= `(7 xx 9)/(3 xx 27)` 

= `(7 xx 3^2)/3^4` 

= `7/9`

So `log (35/33) - log (135/99)`

= `log (7/9)`

 = log 7 – log 9

= log 7 – 2 log 3

3. Add the remaining term: 

`(log 7 - 2 log 3) + log (24/7)`

= `(log 7 + log (24/7)) - 2 log 3`

= `log (7 xx (24/7)) - 2 log 3` 

= log 24 – 2 log 3

4. Express log 24 in terms of log 2 and log 3:

24 = 2³ × 3

So log 24 = 3 log 2 + log 3. 

Therefore,

log 24 – 2 log 3 

= (3 log 2 + log 3) – 2 log 3 

= 3 log 2 – log 3

Hence, `log (35/33) - log (135/99) + log (24/7) = 3 log 2 - log 3`.