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 Pre-Calculus and Calculus
 find value of x in log equation
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alexdc
Junior Member

USA
5 Posts

Posted - 07/30/2011 :  14:28:20  Show Profile  Reply with Quote
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.
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royhaas
Moderator

USA
3059 Posts

Posted - 08/01/2011 :  07:55:01  Show Profile  Reply with Quote
Start by exponentiating each side. For example, if
u = v, then 2^u = 2^v.
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alexdc
Junior Member

USA
5 Posts

Posted - 08/01/2011 :  16:39:51  Show Profile  Reply with Quote
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
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royhaas
Moderator

USA
3059 Posts

Posted - 08/02/2011 :  07:28:22  Show Profile  Reply with Quote
Write the base 2 log in terms of the base 3 log, or vice versa.
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Subhotosh Khan
Advanced Member

USA
9117 Posts

Posted - 08/02/2011 :  15:25:13  Show Profile  Reply with Quote
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.

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alexdc
Junior Member

USA
5 Posts

Posted - 08/02/2011 :  16:22:18  Show Profile  Reply with Quote
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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