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

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So it should be 4^2 = (5-x)/(3-x). In summary, the given equation can be rewritten as (3-x)/(5-x) = 4^2 and using the log rule log_a(b)= c is equivalent to b= a^c, we can solve for x by taking the inverse log of both sides. Some confusion arose regarding the placement of the terms in the numerator and denominator, but it was resolved and the correct equation was determined to be 4^2 = (5-x)/(3-x).
  • #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?
 
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  • #2


What do you know about [tex]a^\log_a x[/tex]?
 
  • #3


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 ?
 
  • #4


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


tweety1234 said:

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


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]
 
  • #7


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


Yeah you're right
 

1. What does the equation log_4(5-x)-log_4(3-x) = 2 represent?

The equation represents a logarithmic function with base 4, where the difference between the inputs of the two logarithms is equal to 2.

2. How do I solve the equation log_4(5-x)-log_4(3-x) = 2?

To solve this equation, you can use the quotient property of logarithms, which states that log(a/b) = log(a) - log(b). You can also use algebraic manipulation to isolate the variable and solve for it.

3. What are the steps to solving log_4(5-x)-log_4(3-x) = 2?

The steps to solving this equation are:

  1. Apply the quotient property of logarithms to rewrite the equation as log_4((5-x)/(3-x)) = 2
  2. Rewrite 2 as log_4(16) since 4^2 = 16
  3. Set the argument of the logarithm equal to the base, in this case 4.
  4. Solve for x.
  5. Check your solution by plugging it back into the original equation.

4. Can I use a calculator to solve this equation?

Yes, you can use a calculator to solve this equation. However, it is important to make sure your calculator is set to the correct base, in this case base 4, and to use parentheses to group the terms correctly.

5. Is there a shortcut to solving this type of equation?

Yes, there is a shortcut known as the logarithmic identities, which states that log(a) - log(b) = log(a/b) and log(a) + log(b) = log(ab). These identities can help simplify the equation and make it easier to solve.

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