I am trying to proof the following expression (without a calculator of course). log_(12)18xxlog_(24)54+5(log_(12)18−log(24)54)=1 I know this isn't a difficult task but it's just killing me. I have tried many things, among which was base transformation to 12 and expressing every logarithm in terms of log_(12)3 and log_(12)2 but every time I try to do it, I mess up something. I don't know if my concentration is terrible or I'm doing something wrong.

Cyrus Travis

Cyrus Travis

Answered question

2022-10-22

Proof the expession log 12 18 × log 24 54 + 5 ( log 12 18 log 24 54 ) = 1
I am trying to proof the following expression (without a calculator of course).
log 12 18 × log 24 54 + 5 ( log 12 18 log 24 54 ) = 1
I know this isn't a difficult task but it's just killing me. I have tried many things, among which was base transformation to 12 and expressing every logarithm in terms of log 12 3 and log 12 2 but every time I try to do it, I mess up something. I don't know if my concentration is terrible or I'm doing something wrong.
Thanks ;) ( if there are more levels to this task, I'd like a hint, not a complete solution)

Answer & Explanation

honotMornne

honotMornne

Beginner2022-10-23Added 12 answers

Let log 12 18 = a , log 24 54 = b. So you want to prove that a b + 5 ( a b ) = 1
Then you get 12 a = 18 and 24 b = 54. Now factor everything in powers of 2 and 3 to get
2 2 a 1 = 3 2 a 2 3 b 1 = 3 3 b .
From this taking log base 2 you will get:
2 a 1 = ( 2 a ) log 2 3 3 b 1 = ( 3 b ) log 2 3.
Furthermore you get
2 a 1 2 a = 3 b 1 3 b .
Now simplify this and see you will get the expression written on the first line.
Sonia Elliott

Sonia Elliott

Beginner2022-10-24Added 4 answers

The follow-your-nose approach would be, in my opinion,
log 12 18 log 24 54 + 5 ( log 12 18 log 24 54 ) = log 18 log 12 log 54 log 24 + 5 ( log 18 log 12 log 54 log 24 ) = log 18 log 12 log 54 log 24 + 5 ( log 18 log 24 log 54 log 12 log 12 log 24 ) = log 18 log 54 + 5 log 18 log 24 5 log 54 log 12 log 12 log 24 .
To save space, write t = log 2 and h = log 3. Then log 18 = log 2 + 2 log 3 = t + 2 h, etc., and we get
log 18 log 54 + 5 log 18 log 24 5 log 54 log 12 log 12 log 24 = ( t + 2 h ) ( t + 3 h ) + 5 ( t + 2 h ) ( 3 t + h ) 5 ( t + 3 h ) ( 2 t + h ) ( 2 t + h ) ( 3 t + h ) = t 2 + 5 t h + 6 h 2 + 15 t 2 + 35 t h + 10 h 2 ( 10 t 2 + 35 t h + 15 h 2 ) 6 t 2 + 5 t h + h 2 = 1.

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