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Numerator log help

  • Thread starter tweety1234
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  • #1
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Homework Statement



[tex] log_4(5-x)-log_4(3-x) = 2 [/tex]



The Attempt at a Solution



[tex] log_4(5-x)-log_4(3-x) = 2 [/tex]
[tex] log_4\frac{3-x}{5-x} = 2 [/tex]

where do I go from here?
 

Answers and Replies

  • #2
Cyosis
Homework Helper
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What do you know about [tex]a^\log_a x[/tex]?
 
  • #3
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What do you know about [tex]a^\log_a x[/tex]?
Dont know, not come across that expression before. Could you please explain.

Thanks.

Are you using the power rule ?
 
  • #4
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It's actually much simpler than that cyosis.

You're on the right track but what do you know about logs?

Consider this;

log 2 (4) = 2 Am I right? Isn't this just another way of saying, 2^2 = 4?

Ok, let's apply this logic to your question.

log 4 (3-x / 5-x ) = 2

Now do the same
 
Last edited:
  • #5
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement



[tex] log_4(5-x)-log_4(3-x) = 2 [/tex]



The Attempt at a Solution



[tex] log_4(5-x)-log_4(3-x) = 2 [/tex]
[tex] log_4\frac{3-x}{5-x} = 2 [/tex]

where do I go from here?
[itex]log_a(b)= c[/itex] is equivalent to [itex]b= a^c[/itex]. Here, a= 4, b= (3-x)/(5- x), and c= 2.
 
  • #6
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[itex]log_a(b)= c[/itex] is equivalent to [itex]b= a^c[/itex]. Here, a= 4, b= (3-x)/(5- x), and c= 2.

Oh right I get it now,
thanks

[tex] \frac{3-x}{5-x} = 4^2 [/tex]
 
  • #7
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Shouldn't the 5-x term be in the numerator and the 3-x term in the denominator?
I'm not 100% certain, but I thought log b - log a = log (b/a), not log (a/b)?
 
  • #8
370
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Yeah you're right
 

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