Logarithms = ???

For the discussion of math. Duh.

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Strychnos
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Logarithms = ???

Postby Strychnos » Thu Sep 13, 2007 8:56 pm UTC

So I'm taking AP BC Calc online at school, which is sort of hard to learn without a physical body there to ask questions and all that.

We're still in the phase of reviewing Trig/AB material and I'm already struggling - I dunno if I'm going to stick with the class, but anyways.

How would you go about solving this:

Image

Solving for x.

I really have no idea how to go about doing this, any pointers would be awesome<3
Image

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antonfire
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Postby antonfire » Thu Sep 13, 2007 9:00 pm UTC

What's the formula about logarithms that allows you to change the base of the log?
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Pesto
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Postby Pesto » Thu Sep 13, 2007 9:01 pm UTC

Can your repost that in text? The image isn't showing up for me.

The image is probably linked correctly, but I'm at work, and they block a lot of things.

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Strychnos
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Postby Strychnos » Thu Sep 13, 2007 9:06 pm UTC

Mm quick replies<3

As far as I know you can write it out as
Image
log(9x^2)/log(4) = log(5x-8 )/log(2)

using log[baseb](a) = loga/logb.

In ugly text the original equation was
log[base4](9x^2) = log[base2](5x - 8 )
Image

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Pesto
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Postby Pesto » Thu Sep 13, 2007 9:14 pm UTC

Well, log(4) and log(2) are just constants. There's a good way to simplify them.

Try and find it.

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Strychnos
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Postby Strychnos » Thu Sep 13, 2007 9:18 pm UTC

Is there any way to simplify them down to something more tangible other than using my calculator?
log(4)=.60206
log(2)=.30103

I'm terribly rusty with this whole process but hopefully it'll all come back soon enough.

Edit:
Just kidding: I just actually looked at the decimals and realized log(4)/log(2) = 2. Lemme try this again.

Edit #2:
Ok, I got x=-2, but when I find the intercept with my calculator, it's at x=4.


Here's my work:
Image


Edit #3:
I think I got it. Instead of making it 2log(9x) I left it as log(9x^2) and turned the other side into log(5x-8 )^2. Cancel the log's, square root both sides to get 3x=5x-8, simplify, x=4.

Also: I appreciate the help :)
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Postby aguacate » Thu Sep 13, 2007 10:00 pm UTC

When you took the square out of the log in the left hand side, you should have taken the square root of everything, not just x^2. I.e., it should have been 3x, not 9x.

edit: one more clarification just to make sure you follow - log(9x^2) = log[(3x)^2] = 2log(3x)
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Strychnos
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Postby Strychnos » Thu Sep 13, 2007 10:08 pm UTC

Aie, thanks.

This is getting easier and easier.

I'm on to expanding/compressing log/ln's now and was wondering if there's any way to expand ln(x^3 - 4) any more than it already is.

I don't think there is, because as far as I know it's only considered expanding if there are two factors being multiplied/divided.
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Postby Cosmologicon » Thu Sep 13, 2007 10:37 pm UTC

Well, you could factor x^3 - 4 into two factors. It would look kind of ugly, but it would be analytic at least. One factor is (x - 4^(1/3)).

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Pesto
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Postby Pesto » Thu Sep 13, 2007 11:30 pm UTC

Looking at the cheat sheet there is an identity for addition. I hadn't seen it before, but that's not saying much.

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Postby Tac-Tics » Thu Sep 13, 2007 11:30 pm UTC

Logs with mixed bases can get confusing, but the really cool thing about logs (or exponents) is there's only one of them! The rest are just constant multiples of each other!

When you see log_b(n), think of it as just notation for ln(n) / ln(b).

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Re: Logarithms = ???

Postby jtniehof » Thu Sep 13, 2007 11:42 pm UTC

You don't always need to change to base e. This problem's simpler if you don't:

Code: Select all

Simplify LHS:
log_4(9x^2) = log_2 ((3x)^2) / log_2 (4)
= (1/2)*2 log_2 (3x)
Then log_2 (3x) = log_2 (5x-8)
3x=5x-8
x=4

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Nath
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Re: Logarithms = ???

Postby Nath » Thu Sep 13, 2007 11:46 pm UTC

Strychnos wrote:Image

Solving for x.


log_4(9 x^2) = log_2(5x - 8)

log_4(9 x^2) = log_2(9 x^2) * log_4(2) = log_2(9 x^2) * 1/2

log_2(9 x^2) = 2 * log_2(5x - 8) = log_2( (5x - 8)^2 )

(3x)^2 = (5x-8)^2

3x = (+/-) (5x - 8)

x = 1 or 4.

I think.

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Postby gmedina » Thu Sep 13, 2007 11:56 pm UTC

Nath wrote:x = 1 or 4

No. x cannot be equal to 1; otherwise log_2(5x-8) would yield log_2(-3).
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Nath
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Postby Nath » Fri Sep 14, 2007 12:01 am UTC

gmedina wrote:
Nath wrote:x = 1 or 4

No. x cannot be equal to 1; otherwise log_2(5x-8) would yield log_2(-3).


Doh. Yes.

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Postby Mathmagic » Fri Sep 14, 2007 1:34 am UTC

Pesto wrote:Looking at the cheat sheet there is an identity for addition. I hadn't seen it before, but that's not saying much.

With the addition identity, they just factored out 'a' from inside the logarithm, then used the division identity on the 'c/a' left over in the log(1 +/- c/a) logarithm, and then for some expressed it as an exponential with base b...
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Postby imMAW » Fri Sep 14, 2007 3:42 am UTC

Ok, im not sure if somebodies already done this, but its late at night and i dont want to read all that.
log4 9x^2 = log2 5x-8 is the problem
with problems like this, the easiest thing to check for first is if you can get both sides in the same base. In this case you can - taking the square root of both the base and argument of a log will not change the number (neither will taking the cube root, or squaring it, etc). So, taking the square root of the left, we get
log2 3x = log2 5x-8
now if logx a = logx b, then a = b. So;
3x = 5x-8
Let's hope you can solve that, if you cant it's my opinion that you should drop calc BC :)

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Pathway
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Postby Pathway » Fri Sep 14, 2007 9:50 pm UTC

Just think of what a logarithm represents.

So:

log4 9x^2 = log2 5x-8

9x^2 = 4^(log2 5x-8)

because

Highlight for the key to all log problems wrote:Log is the inverse of an exponential.


Notice that the only remaining base is 2, and 4 is a power of 2.

So since 4 = 2^2, we have

9x^2 = (2^2)^(log2 5x-8)

and by rules for exponents, you know it's the same as

9x^2 = 2^(2*(log2 5x-8))

Now a good way to solve this problem might be to take a square root of both sides to get 3x = 2^(log2 5x-8). But that's no fun.

What is fun is the following:

9x^2 = 2^(log2 ((5x-8)^2))

Now 2^log2(something inside) = something inside, since

Highlight for the key to all log problems wrote:Log is the inverse of an exponential.


So that gets us to:

9x^2 = (5x-8)^2

Now it's profitable to note that 9 is a perfect square--that is, 3^2 = 9.

So (3x)^2 = (5x-8)^2

So 3x = 5x-8,

2x = 8,

x = 4.
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Postby gmedina » Fri Sep 14, 2007 10:15 pm UTC

Pathway wrote:So (3x)^2 = (5x-8)^2

So 3x = 5x-8,

2x = 8,

x = 4.

No. a^2=b^2 doesn't imply a=b. Anyway, what's the point answering over and over (and, in some cases, with errors) a problem when all that was needed was provided by antonfire in the first reply?
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Pathway
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Postby Pathway » Fri Sep 14, 2007 11:34 pm UTC

In the other case, x=1 implies that there exists log2 (-3).
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Postby gmedina » Sat Sep 15, 2007 2:16 am UTC

Pathway wrote:In the other case, x=1 implies that there exists log2 (-3).

A little too late to add an explanation, in my opinion. (Read 4 or 5 replies above.)
Edit:
Pathway wrote:...x=1 implies that there exists log2 (-3)

Again, no.
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Postby Govalant » Mon Sep 17, 2007 2:53 am UTC

In this case it's a bit easier, because you just raise the base and x to the same number so:

Code: Select all

log4(9x^2) = log2(5x-8)

log2(sqrt(9x^2)) = log2(3x) = log2(5x-8)

3x = 5x-8
2x = 8
x = 4


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