Basic Log Problem..

MiraKul

Malaysia498 Posts

April 16 2011 17:01 GMT

The time is 12.50am. And I've been grinding all my brain juice out to solve 1 simple logarithm problem. I dont know what went wrong.. Need someone to help me out pls.

Prenotes: (*) <-- number in bracket is the base
1) Given log(2) 3 =1.585 , log(2) 5 = 2.322 without using calculator, evaluate log(2) 60.

So, I derived the log(2) 60 = log(2) [(3x5) + (3x5) + (3x5) + (3x5)]
= [log(2) 3 + log(2) 5] x 4
= [(1.585 + 2.322) x 4]
= 15.628

But, when I rechecked from my calculator, the ans is 5.906.

I couldnt get it.. I hope TL members can help me.

Thanks in advance. Ima head back to the prob.

Got a pm from a Good samaritan. : "i dont think hw help threads are allowed"

oops. im sorry.. didnt know that. Mod pls close if it's not allowed..

djcube

United States985 Posts

April 16 2011 17:10 GMT

lg(60) = lg(2)+ lg(2)+lg(3)+lg(5)

= 1+1+lg(3) + lg(5)
= 5.906

MisterD

Germany1338 Posts

April 16 2011 17:16 GMT

so here's what you did:

log_2(60) = log_2(3*5 + 3*5 + 3*5 + 3*5) = log_2(3*5*4)
so far so good. but - then you did:
= log_2(3*5*4) = 4*log_2(3*5)
which is wrong! the correct thing would have been:
log_2(3*5*4) = log_2(4) + log_2(3*5)
then you could express 4=2*2 from which you get to:
= log_2(2*2) + log_2(3*5) = log_2(2) + log_2(2) + log_2(3) + log_2(5)
which is where first reply started off.

DarkPlasmaBall

United States45089 Posts

April 16 2011 18:45 GMT

On April 17 2011 02:10 djcube wrote:
lg(60) = lg(2)+ lg(2)+lg(3)+lg(5)

= 1+1+lg(3) + lg(5)
= 5.906

This is correct.

You can split up 60 into its factors and evaluate the sum of each of those logs, as he showed above. 60 = 2*2*3*5, and so you can do log(2) + log(2) + log(3) + log(5) to calculate log(60).

XXGeneration

United States625 Posts

April 16 2011 21:02 GMT

When you multiply the sum of the logs by 4, you're actually increasing the power of the parenthesised portion by 4.

Ex. 4 log(2) 5 = log(2) 5^4 = log(2) 625.

This could explain why you were wrong.

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