Quick Log Question

[sysD]

Ok, so I'm a bit stuck.

 

Here's the problem:

 

 

 

log(base9)(x-6) + log(base9)(x+2) = 1

 

Here's my work.

 

At first i simply tried to solve by rooting:

(x-6)(x+2) = 0

x = 6 or x = -2

But realized this would not work, as any rooting would result in having one of the logarithms above yield an invalid result (eg. log_9(0) )

 

I've tried another way (isolating x), but I'm stuck.

 

(x-6)(x+2) = 0

x^2 + 2x - 6x - 12 = 0

x^2 -4x = 12

 

The only thing I can think of doing at this point is

 

x - 2(x^(0.5)) = 12^(0.5)

 

And that's kind of useless (i think).

Any ideas?

Edited March 8, 2011 by sysD

[mooeypoo]

What are the rules of logs? Do you remember? What can you do with Log(whatever) + Log(whatever else) ?

[Fuzzwood]

To help out a bit more: how would you turn that = 1 to something that is also log(base 9)? And yes, you still need to root after that. log(base9)(x-6) + log(base9)(x+2) = log(base9)[(x-6)*(x+2)]

Edited March 8, 2011 by Fuzzwood

[sysD]

As far as I remember, Log(baseY)x + log(baseY)2x = log(baseY)(x(2x))

 

But I've already done this:

 

log(base9)(x-6) + log(base9)(x+2) = 1

log_9((x-6)(x+2)) = 1

log_9((x-6)(x+2)) = log_9(0)

log_9(x^2 - 4x - 12) = log_9(0)

x^2 - 4x = 12

I'm not sure about this next part.

 

2logx - log4x = log12

2logx = log(12) + log(4x)

2logx = log(12)(4x)

2logx / log4x = log12

log12 = 2logx / log4x

log12 = log((x^2)-(4x))

I'm going in circles here.

 

12 = x^2 - 4x

 

I'm still stuck. Could I have another hint please?

 

To help out a bit more: how would you turn that = 1 to something that is also log(base 9)? And yes, you still need to root after that. log(base9)(x-6) + log(base9)(x+2) = log(base9)[(x-6)*(x+2)]

 

 

Sorry, I still don't get it.

 

If I was to turn it into a logarithm with a base of 9, it would be:

 

log_9(1)=0

The only number I can think of that'll apply the Log Rule of Multiplication is 1.

log_9(1) + log_9(1) = 0

log_9(1)(1) = 0

 

Seems kind of useless.. I'm missing something here.

[mooeypoo]

Okay, so you have this:

[math]x^2 - 4x = 12[/math]

 

You don't need to go back to having logs or powers anymore, this is just remembering your albebra. Try putting this in a form where it might factor? If it can't be easily factored, can you think of ways to solve that type of equation? Perhaps some roots related formula?

 

Sorry, I still don't get it.

 

If I was to turn it into a logarithm with a base of 9, it would be:

 

log_9(1)=0

The only number I can think of that'll apply the Log Rule of Multiplication is 1.

log_9(1) + log_9(1) = 0

log_9(1)(1) = 0

 

Seems kind of useless.. I'm missing something here.

 

You already did this by rewriting the right side of the equation to log_9(0).

 

I'm sorry, you just seemed to have skipped a few steps so we were trying to lead you to continue from where you stopped. From your second reply, it seems you already managed to get rid of the logs (which is good) and now you just need to continue and solve for x.

 

You're in the right track, just keep going.

[sysD]

Ahhhh, the quadratic formula:

 

x = -b +- ( (b^2 - 4ac)^(.5) )/2a

 

I always seem to overlook this lil equation because its just so darned simple to use.

I immediately go for the hardest solutions XD

 

Anyways, onwards:

 

x = 4 +- ( (16 + 48)^(.5) ) /2

x = 4 +- ( (64)^(.5) ) /2

x = 4 +- 8 / 2

x = 4 +- 4

 

x = 8

 

OR

 

x = 0 (undefined)

 

Therefore, x = 8

 

Ugh, ok I just plugged that Into the original formula and it is wrong....

 

x=8 yields ~1.363416514 instead of ~1

 

and

 

in

x^2 - 4x -12 = 0

64 - 32 - 12 = 0

32 - 12 = 0

20 = 0

 

As you might already know, twenty does NOT equal zero. Something is very wrong.

 

 

I can't seem to figure out any way to factor this: x^2 - 4x -12 = 0

I don't think it can be factored in a useful way, can it?

Edited March 9, 2011 by sysD

[mooeypoo]

This is the quadratic formula:

 

[math]\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/math]

 

You wrote it slightly differently, which changed your answer (and made it wrong ). Try again.

[sysD]

Hey, thanks money, that was a stupid error, haha. The thing is, I see now where I skipped a step, and I'm not sure how it led me to the right answer.

 

Well, the answer's correct now, but I see what you mean about skipping steps. Let's go back to the beginning, so that I actually completely understand what I am doing.

 

We have:

 

log_9(x-6) + log_9(x+2) = 1

 

The next step is to convert the 1 into something with the base 9 that results in a value of "1"

Hmm, what could that be?

 

log_9(x-6) + log_9(x+2) = 9^0

 

And in log form..

 

log_9(x-6) + log_9(x+2) = log_9(1)

 

Let's apply the laws of logs.

 

log_9( (x-6)(x+2) ) = log_9(1)

 

 

 

Here's where I'm stuck, again.

 

I want to get to: (x-6)(x+2) = 0

 

But doing so would require me to solve the log on one side of the equation, while dropping the log on the other side. Which doesn't seem right.

 

What seems right is:

log_9( (x-6)(x+2) ) = log_9(1)

(x-6)(x+2) = 1

 

This seems correct because in the form "log_b(y) = z," the "log_b" is dropped, and they "y" value is kept

 

Which goes to:

 

x^2 - 4x -12 -1 = 0

x^2 - 4x - 13 = 0

 

And in the quadratic equation...

 

x = (-b +- (b^2 - 4ac)^(.5)) / 2a

 

x = (4 +- (16 - 52)^(.5)) / 2a

 

x = (4 +- 6i) / 2

 

x = 2 +- 3i

 

 

So in this method:

 

x = 2 + 3i

 

OR

 

x = 2 - 3i

 

 

I'm not sure how to check if these are correct answers, since they are complex numbers, which I have little experience with [other than knowing "i=(-1)^(.5)"].

Edited March 9, 2011 by sysD

[mooeypoo]

I have to run in a second, but when I did this with the previously given x^2-4x-12=0 I got 6 and -2 as my values for x. Since your "a" is positive and "b" and "c" are negative in this equation, the inside of the root in the quadratic formula should be positive.. +4ac, not -4ac.. that would prevent the complex number.

 

 

Verify this... a complex number here seems odd.

 

 

 

 

Okay ,I see what's going on.. you're rushing again. Which is ironic, since I'm the one in a rush this morning.

 

Look here:

We have:

 

log_9(x-6) + log_9(x+2) = 1

 

The next step is to convert the 1 into something with the base 9 that results in a value of "1"

Hmm, what could that be?

 

log_9(x-6) + log_9(x+2) = 9^0

Great so far, but then look how you write your log:

 

And in log form..

 

log_9(x-6) + log_9(x+2) = log_9(1)

No. In log form, the right hand side is log_9(0), like you wrote above in the first posts.

meh. Sorry, morning and I'm rushing it too. To get log(something)=1, you need to have log_9(9), not log_9(0) like I wrote before which is undefined. Apologies. Still, don't rush it. The quadratic root should be positive.

 

Don't rush.....

 

Slow down, you're in the right track, your errors are just concentration and rushing it. Write this properly, and then solve the quadratic equation properly and pay attention to the signs.

 

It has two solutions.

 

~mooey

[Fuzzwood]

When you have

log_9(x-6) + log_9(x+2) = log_9(1), you can kick the logs away so you remain with (x-6)(x+2) = 9

[sysD]

To get log(something)=1, you need to have log_9(9), not log_9(0) like I wrote before which is undefined. Apologies. Still, don't rush it. The quadratic root should be positive.

 

 

Ah, ok. I was looking at this the wrong way.

I was using "1" as the "x" value in log_b(x) = y

It should have been the "y" value, as a log equation would result in "1"

So in:

log_9(x) = 1

9^1=x

x=9

 

Starting Over:

 

log_9((x-6)(x+2)) = 1

log_9((x-6)(x+2)) = log_9(9)

(x-6)(x+2) = 9

x^2 - 4x - 12 - 9 = 0

x^2 - 4x - 21 = 0

 

Simple trinomial factoring:

 

(x-7)(x+3) = 0

 

x= 7

OR

x= -3 (undefined)

 

Therefore, x = 7

 

 

 

Checking Answer:

 

log_9(7-6) + log_9(7+2) = 1

log_9(1) + log_9(9) = 1

log_9(1)(9) = 1

log_9(9) = 1

In exponential form:

9^1 = 9

 

 

 

Sweet.

Nothing feels as good as FINALLY solving a math problem. xp+10 (chugs beer)

 

Thanks for the help, Mooey.

Haha, I just realized I called you "money" before... just another example of me rushing over things and my brain filling in the gaps.

I have trouble taking things slow, which is why I take the pains to check my answers so many times. Its saved me more than once.

Edited March 9, 2011 by sysD

[mooeypoo]

Hehe apparently I rushed too, since I didn't even notice its okay.

 

One tip I can give you though if you know you are rushing, other than checking the answers (which is awesome) is to make a mental list of potential slipups.

 

I rush too in exams, which makes it extremely annoying and frustrating to get a lower grade later. What I try to do is remember the things I slip on. For me, its mostly signs that I need to be careful of and translations regarding exponents. So I take extra seconds to verify these.

 

Also, if I get stuck on one question over and over, instead of trying to figure out what I did wrong (which is tough sometimes, because you tend to stick to the same methodology and miss errors) I rip out the page, do another question to clear my mind off this one, and come back to doing the problematic one from scratch.

 

I think of it as closing all the windows and resetting my brain. Maybe one day I will switch to mental linux

 

Cheers and good luck!

 

~mooey