Author 
Topic 

alexdc
Junior Member
USA
5 Posts 
Posted  07/30/2011 : 14:28:20

I'm not understanding where I'm going wrong on this one, but here it is:
find x when
log (sub 2) x = log (sub 3) x + 1
Answer:
x = 2^( 1/(1log (sub2) 3) ) or x = 3^(1/(log (sub 2) 3  1)
My work:
start by dividing each side by log (sub 2) x to get:
1 = ( (log (sub 3) x) / (log (sub 2) x) ) + ( 1 / (log (sub 2) x) )
then, use the identity:
( (log (sub a) y) / (log (sub b) y) ) = log (sub a) b
A bit of algebra, and this identity is also:
1 / (log (sub b) y) = (log (sub a) b) * (log (sub a) y)
Applying this identity to the problem, we have:
y = x a = 3 b = 2
So that:
( (log (sub 3) x) / (log (sub 2) x) ) = log (sub 3) 2
and:
( 1 / (log (sub 2) x) ) = (log (sub 3) 2) * (log (sub 3) x)
so our equation is now:
1 = ( log (sub 3) 2 ) + ( (log (sub 3) 2) * (log (sub 3) x) )
Divide both sides by ( log (sub 3) 2 ) to get:
1 / ( log (sub 3) 2 ) = 1 + ( log (sub 3) x )
subtract both sides by 1 to get:
( 1 / ( log (sub 3) 2 ) )  1 = ( log (sub 3) x )
we know that log (sub a) b = c is the same as b = a ^ c, so:
x = 3 ^ ( ( 1 / ( log (sub 3) 2 ) )  1 )
which is not one of the answers. Similarly, by starting to solve this by dividing the whole equation ( log (sub 3) x ), I get the answer:
2^( 1/(1log (sub3) 2) )
I'm not seeing what I'm doing wrong here, but my answers don't check, and answers in the book do.
Any help is appreciated! Thank you. 


royhaas
Moderator
USA
3059 Posts 
Posted  08/01/2011 : 07:55:01

Start by exponentiating each side. For example, if u = v, then 2^u = 2^v. 


alexdc
Junior Member
USA
5 Posts 
Posted  08/01/2011 : 16:39:51

I think I understand what you are saying, but I'm not seeing the next step. I take it the idea would be to make each side as a power of 2 or 3, and get either (making both sides a power of 2):
x = 2^( log (sub 3) x + 1 )
simplified:
x = 2^( log (sub 3) x ) * (2^1)
simplified:
x = 2^( log (sub 3) x ) * 2
or (making both sides a power of 3):
3^( log (sub 2) x ) = 3^( log (sub 3) x + 1 )
simplified:
3^( log (sub 2) x ) = 3^( log (sub 3) x ) * (3^1)
simplified:
3^( log (sub 2) x ) = 3x
However, in both cases, I'm not sure what to do after the above listed steps. I have a feeling I'm missing something incredibly obvious here. 
Edited by  alexdc on 08/01/2011 16:40:39 


royhaas
Moderator
USA
3059 Posts 
Posted  08/02/2011 : 07:28:22

Write the base 2 log in terms of the base 3 log, or vice versa. 


Subhotosh Khan
Advanced Member
USA
9117 Posts 
Posted  08/02/2011 : 15:25:13

quote: Originally posted by alexdc
I'm not understanding where I'm going wrong on this one, but here it is:
find x when
log (sub 2) x = log (sub 3) x + 1
log_{3}x = log_{3}2 * log_{2}x
Answer:
x = 2^( 1/(1log (sub2) 3) ) or x = 3^(1/(log (sub 2) 3  1)
My work:
start by dividing each side by log (sub 2) x to get:
1 = ( (log (sub 3) x) / (log (sub 2) x) ) + ( 1 / (log (sub 2) x) )
then, use the identity:
( (log (sub a) y) / (log (sub b) y) ) = log (sub a) b
A bit of algebra, and this identity is also:
1 / (log (sub b) y) = (log (sub a) b) * (log (sub a) y)
Applying this identity to the problem, we have:
y = x a = 3 b = 2
So that:
( (log (sub 3) x) / (log (sub 2) x) ) = log (sub 3) 2
and:
( 1 / (log (sub 2) x) ) = (log (sub 3) 2) * (log (sub 3) x)
so our equation is now:
1 = ( log (sub 3) 2 ) + ( (log (sub 3) 2) * (log (sub 3) x) )
Divide both sides by ( log (sub 3) 2 ) to get:
1 / ( log (sub 3) 2 ) = 1 + ( log (sub 3) x )
subtract both sides by 1 to get:
( 1 / ( log (sub 3) 2 ) )  1 = ( log (sub 3) x )
we know that log (sub a) b = c is the same as b = a ^ c, so:
x = 3 ^ ( ( 1 / ( log (sub 3) 2 ) )  1 )
which is not one of the answers. Similarly, by starting to solve this by dividing the whole equation ( log (sub 3) x ), I get the answer:
2^( 1/(1log (sub3) 2) )
I'm not seeing what I'm doing wrong here, but my answers don't check, and answers in the book do.
Any help is appreciated! Thank you.



alexdc
Junior Member
USA
5 Posts 
Posted  08/02/2011 : 16:22:18

Just noticed that. I screwed up basic algebra, that was my mistake. This:
( (log (sub a) y) / (log (sub b) y) ) = log (sub a) b
DOES NOT equal this:
1 / (log (sub b) y) = (log (sub a) b) * (log (sub a) y)
After ANY amount of algebra. I multiplied both sides by (log (sub a) y), instead of dividing. Stupid mistake, and the answer is correct after using the proper identity.
Thanks for the help! 



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