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Pre-Calculus and Calculus

find value of x in log equation

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[quote][i]Originally posted by alexdc[/i]
[br]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/(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) 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/(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.
[/quote]

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T O P I C R E V I E W
alexdc	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/(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) 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/(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.
5 L A T E S T R E P L I E S (Newest First)
alexdc	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!
Subhotosh Khan	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 log₃x = log₃2 * log₂x Answer: 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) 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/(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	Posted - 08/02/2011 : 07:28:22 Write the base 2 log in terms of the base 3 log, or vice versa.
alexdc	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.
royhaas	Posted - 08/01/2011 : 07:55:01 Start by exponentiating each side. For example, if u = v, then 2^u = 2^v.