Homework Help: Numerator log help

tweety1234 · Apr 29, 2009

1. The problem statement, all variables and given/known data

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

3. 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?

Cyosis · Apr 29, 2009

Re: logarithms

What do you know about [tex]a^\log_a x[/tex]?

tweety1234 · Apr 29, 2009

Re: logarithms

Cyosis said: ↑

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 ?

Chewy0087 · Apr 29, 2009

Re: logarithms

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

HallsofIvy · Apr 29, 2009

Re: logarithms

tweety1234 said: ↑

1. The problem statement, all variables and given/known data

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

3. 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.

tweety1234 · Apr 29, 2009

Re: logarithms

HallsofIvy said: ↑

[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]

DorianG · May 1, 2009

Re: logarithms

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)?

Chewy0087 · May 1, 2009

Re: logarithms

Yeah you're right