- #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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- Thread starter tweety1234
- Start date

- #1

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

[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?

- #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

- 370

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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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[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.## 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?

- #6

- 112

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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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