Solving Logarithmic Equation: Why Isn't It Working?

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The discussion revolves around solving the logarithmic equation log_2 x + log_4 x = 2. The confusion arises from the application of the change of base rule and how to combine logarithms effectively. The correct approach involves expressing log_4 x in terms of log_2 x, leading to the equation log_2 x + (log_2 x)/2 = 2. This simplifies to 1.5 * log_2 x = 2, clarifying how the factor of 3/2 is derived. Ultimately, the method discussed is valid, and understanding the manipulation of logarithmic bases is key to solving the equation correctly.
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Homework Statement



I can't seem to get my head round this problem, I know how to use the 'change the base rule'

log_2 x + log_4x = 2

\frac{logx}{log2} + \frac{logx}{log4} = 2

why is this not correct ??
 
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It looks correct - what's the problem
(of course everytime I stray into maths I make an dumb mistake!)
 
mgb_phys said:
It looks correct - what's the problem
(of course everytime I stray into maths I make an dumb mistake!)

well the answer book gives the solution like this,

log_2x + \frac{log_2 x}{log_2 4} = 2 don't understand why its base 2 here?

log_2 x + \frac{log_2 x}{2} =2

\frac{3}{2}log_2 x = 2 don't understand were 3/2 came from ?

log_2 x = \frac{4}{3}

x = 2^{\frac{4}{3} }
 
They are just collapsing it all into the same base log.

log_2 x + \frac{log_2 x}{2} =2 is just

1.5 * log_2 x =2 which is

\frac{3}{2} * log_2 x =2

It's the same answer as you get - but you can do it this way without needing to calculate log() of anything
 
tweety1234 said:
log_2 x + log_4x = 2

\frac{logx}{log2} + \frac{logx}{log4} = 2

so is this method still correct? can I just add both logs up ?
 
Except you are trying to find X so first you have to multiply out log(2) and log(4)
 
mgb_phys said:
They are just collapsing it all into the same base log.


log_2 x + \frac{log_2 x}{2} =2 is just

1.5 * log_2 x =2 which is

\frac{3}{2} * log_2 x =2

It's the same answer as you get - but you can do it this way without needing to calculate log() of anything

sorry, I still don't get where 1.5 comes from ? I thought you just multiply both sides by two to get rid of the fraction ?

2log_2 x + log_2 x = 4
 
log_2 x + \frac{log_2 x}{2} = 2

Which if you ignore the logs for now is just; a + a/2 = 2

a(1+1/2) = 2

1.5a = 2

3/2 a =2
 
mgb_phys said:
They are just collapsing it all into the same base log.

log_2 x + \frac{log_2 x}{2} =2 is just

1.5 * log_2 x =2 which is

\frac{3}{2} * log_2 x =2

It's the same answer as you get - but you can do it this way without needing to calculate log() of anything

mgb_phys said:
log_2 x + \frac{log_2 x}{2} = 2

Which if you ignore the logs for now is just; a + a/2 = 2

a(1+1/2) = 2

1.5a = 2

3/2 a =2

oh I get it now, thanks a lot for your help!
 
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