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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 whenlog (sub 2) x = log (sub 3) x + 1Answer:x = 2^( 1/(1-log (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) bA 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 = xa = 3b = 2So that:( (log (sub 3) x) / (log (sub 2) x) ) = log (sub 3) 2and:( 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/(1-log (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
3039 Posts

 Posted - 08/01/2011 :  07:55:01 Start by exponentiating each side. For example, ifu = 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 ) * 2or (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 ) = 3xHowever, 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
3039 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

USA
9114 Posts

 Posted - 08/02/2011 :  15:25:13 quote:Originally posted by alexdcI'm not understanding where I'm going wrong on this one, but here it is:find x whenlog (sub 2) x = log (sub 3) x + 1log3x = log32 * log2xAnswer:x = 2^( 1/(1-log (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) bA 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 = xa = 3b = 2So that:( (log (sub 3) x) / (log (sub 2) x) ) = log (sub 3) 2and:( 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/(1-log (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) bDOES 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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