Solving log_4(5-x)-log_4(3-x) = 2: Step-by-Step Guide

[tweety1234]

Homework Statement

$$log_4(5-x)-log_4(3-x) = 2$$

The Attempt at a Solution

$$log_4(5-x)-log_4(3-x) = 2$$
$$log_4\frac{3-x}{5-x} = 2$$

where do I go from here?

 

[Cyosis]

What do you know about $$a^\log_a x$$?

 

[tweety1234]

What do you know about $$a^\log_a x$$?

Dont know, not come across that expression before. Could you please explain.

Thanks.

Are you using the power rule ?

 

[Chewy0087]

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]

Homework Statement

$$log_4(5-x)-log_4(3-x) = 2$$

The Attempt at a Solution

$$log_4(5-x)-log_4(3-x) = 2$$
$$log_4\frac{3-x}{5-x} = 2$$

where do I go from here?

$log_a(b)= c$ is equivalent to $b= a^c$. Here, a= 4, b= (3-x)/(5- x), and c= 2.

 

[tweety1234]

$log_a(b)= c$ is equivalent to $b= a^c$. Here, a= 4, b= (3-x)/(5- x), and c= 2.

Oh right I get it now,
thanks

$$\frac{3-x}{5-x} = 4^2$$

 

[DorianG]

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]

Yeah you're right