## Logarithms = ???

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

Strychnos
Posts: 41
Joined: Mon Oct 23, 2006 10:00 pm UTC

### Logarithms = ???

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:

Solving for x.

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

antonfire
Posts: 1772
Joined: Thu Apr 05, 2007 7:31 pm UTC
What's the formula about logarithms that allows you to change the base of the log?
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Pesto
Posts: 737
Joined: Wed Sep 05, 2007 5:33 pm UTC
Location: Berkeley, CA
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.

Strychnos
Posts: 41
Joined: Mon Oct 23, 2006 10:00 pm UTC
Mm quick replies<3

As far as I know you can write it out as

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 )

Pesto
Posts: 737
Joined: Wed Sep 05, 2007 5:33 pm UTC
Location: Berkeley, CA
Well, log(4) and log(2) are just constants. There's a good way to simplify them.

Try and find it.

Strychnos
Posts: 41
Joined: Mon Oct 23, 2006 10:00 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:

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

aguacate
Posts: 209
Joined: Fri Feb 16, 2007 10:29 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)

Strychnos
Posts: 41
Joined: Mon Oct 23, 2006 10:00 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.

Cosmologicon
Posts: 1806
Joined: Sat Nov 25, 2006 9:47 am UTC
Location: Cambridge MA USA
Contact:
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)).

Pesto
Posts: 737
Joined: Wed Sep 05, 2007 5:33 pm UTC
Location: Berkeley, CA
Looking at the cheat sheet there is an identity for addition. I hadn't seen it before, but that's not saying much.

Tac-Tics
Posts: 536
Joined: Thu Sep 13, 2007 7:58 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).

jtniehof
Posts: 312
Joined: Mon Sep 10, 2007 9:00 pm UTC

### Re: Logarithms = ???

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-8x=4`

Nath
Posts: 3148
Joined: Sat Sep 08, 2007 8:14 pm UTC

### Re: Logarithms = ???

Strychnos wrote:

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.

gmedina
Posts: 109
Joined: Mon Aug 27, 2007 4:36 pm UTC
Location: Colombia
Nath wrote:x = 1 or 4

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

Nath
Posts: 3148
Joined: Sat Sep 08, 2007 8:14 pm 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.

Mathmagic
It's not as cool as that Criss Angel stuff.
Posts: 2926
Joined: Thu Nov 30, 2006 12:48 am UTC
Location: In ur fora posting in teh threads
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...
Axman: That, and have you played DX 10 games? It's like having your corneas swabbed with clits made out of morphine.
Pathway: cocks cocks cocks

imMAW
Posts: 112
Joined: Sun Mar 18, 2007 10:30 pm UTC
Location: Chicago
Contact:
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

Pathway
Leon Sumbitches...?
Posts: 647
Joined: Sun Oct 15, 2006 5:59 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.
SargeZT wrote:Oh dear no, I love penguins. They're my favorite animal ever besides cows.

The reason I would kill penguins would be, no one ever, ever fucking kills penguins.

gmedina
Posts: 109
Joined: Mon Aug 27, 2007 4:36 pm UTC
Location: Colombia
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?
In Memoriam

Pathway
Leon Sumbitches...?
Posts: 647
Joined: Sun Oct 15, 2006 5:59 pm UTC
In the other case, x=1 implies that there exists log2 (-3).
SargeZT wrote:Oh dear no, I love penguins. They're my favorite animal ever besides cows.

The reason I would kill penguins would be, no one ever, ever fucking kills penguins.

gmedina
Posts: 109
Joined: Mon Aug 27, 2007 4:36 pm UTC
Location: Colombia
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.
In Memoriam

Govalant
Posts: 249
Joined: Mon Sep 17, 2007 2:50 am UTC
Location: Rosario, Argentina
Contact:
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-82x = 8x = 4`