Equation problem

Deon588 · 2011-09-23 20:21:55

Hi all. I have no idea if i'm doing this the right way, If I am where do I go from here? Thanks in advance

anonimnystefy · 2011-09-23 22:25:31

hi Deon588

what is the question?

zetafunc. · 2011-09-23 23:16:07

I think he is trying to solve the equation

To solve this, try splitting both the LHS and the RHS into the sum of two logarithms (can you separate x/4 and x/8?). Then you can evaluate some of your new logarithms since some of them won't involve an unknown (x). Then, you just need to use the change of base rule for one of your logarithms and you'll get a nice, easy equation to solve for x. Reply if you want more information.

anonimnystefy · 2011-09-23 23:18:59

hi zetafunc.

i think that's right.

have you thought of becoming a member?

Deon588 · 2011-09-24 00:46:19

Sorry for not being totally clear annonimnystify. The question is write the domain of f(x) and solve the equation. Zetafunc do you mean break it up like this log_2(x-4)=log_4((x-8) sorry for writing it like this, I don't know much about LaTeX yet... If this is the way to go how would the change of base step look? should it be log_2/log_2(x-8)? Thanks a lot

zetafunc. · 2011-09-24 01:38:54

The multiplicative law for logarithms is as follows;

Similarly, the division law for logarithms is;

Bases must be the same for this to work.

So how can we break up

?

Write it like this;

. Can you see where to go from there?

Also, I don't agree that the domain of your function is simply x ∈ R. Can you think of a problem with this?

Deon588 · 2011-09-24 04:52:40

Should I do it like this?

Should the domain be >0? Thanks again.

Bob · 2011-09-24 21:25:46

hi Deon 588

I'm a bit confused about what the question was here, but I'll say what I think it is and then do that.

Question: What is the real number domain restriction?

Subtracting this constant moves the graph down parallel to the y axis but doesn't alter the domain restriction.

The log function is not defined for x ≤ 0 so that deals with that.

Question: Solve this equation:

First get everything in the same log base. You could choose 2 or 4 here, but I think 2 is maginally easier.

but log 4 = 2 so

So either x = 0 which is outside the domain and so not possible or

Bob

Last edited by Bob (2011-09-25 07:16:32)

Deon588 · 2011-09-24 23:06:59

Hi Bob. Thanks a lot that was a very nice detailed explanation. The only part I don't totally understand is where you changed the base

Deon588 · 2011-09-24 23:09:05

Sorry I tried entering the LaTeX. Any idea where my LaTex mistake was?
Thanks

Bob · 2011-09-24 23:53:07

hi Deon588

Answer in prepartion. Stay on-line

Bob

anonimnystefy · 2011-09-24 23:53:56

hi Deon588

try this:

\\ \log_2\frac{x}{4} =\left(\frac{\log_2\frac{x}{8}}{\frac{1}{2}}\right)\\ \mbox{This is the first equation I do which has logarithms and fractions, this is how I remember the change of base formula}\\ \log_2 \frac{x}{4}=\left(\frac{\log_2 \frac{x}{8}}{\log_2 4}\right)\\\mbox{Does that become}\frac{1}{2}\mbox{because it is the demumerator?}

you can use the equation editor here:Equation Editor

bobbym · 2011-09-24 23:54:25

Repaired it.

Bob · 2011-09-25 00:02:14

Hi again,

There is a formula for changing log base, but I've always been hopeless at remembering formulas, so I've always worked it afresh, each time I need it.

Convert to a power expression:

Now put into the log base you actually want:

so

As for the Latex, it is pretty horrid for these expressions.

If you click on the maths expression, you'll see the code minus the square brackets math. That might help you in future.

Bob

Deon588 · 2011-09-25 06:35:24

Thanks a lot for all the answers. Bob when you have

as the denumerator doesn't that become 2 and not

Bob · 2011-09-25 07:04:04

Hi Deon588,

Yes, I've managed to make two errors there, which conveniently cancel out. Whoops.

My rather pathetic excuse is that I had a very late night yesterday, or rather in the early hours of this morning.

Well done for spotting it and for insisting I do it correctly.

Notice the answer was correct, which is why I was happy before, that I was on the right track.

I've gone back and edited post #8 to show what I now hope is the correct version.

Bob

Last edited by Bob (2011-09-25 07:19:25)

reconsideryouranswer · 2011-09-25 15:50:35

Deon588 wrote:

Sorry for not being totally clear annonimnystify.
The question is write the domain of f(x) and solve the equation.
Zetafunc do you mean break it up like this log_2(x-4)=log_4((x-8)
sorry for writing it like this, I don't know much about LaTeX yet...
If this is the way to go how would the change of base step look?
should it be log_2/log_2(x-8)? Thanks a lot

Deon588,

please make some breaks in your lines such as in this amended
quote box above.

Deon588 · 2011-09-28 08:30:28

I had some internet problems so havn't been online. Bob lol no worries and thanks a lot That answer makes a lot of sense I definitely learned something new.

Bob · 2011-09-28 09:45:29

hi Deon588,

Good to hear from you.

Isn't it amazing how we fall apart when our computers let us down.
How did we manage without them?

I'm glad to be making sense at last.

I tried to cancel an x without squaring out first. So I missed the 'over 4 needs to be squared' bit.

And I knew that x = 2 was right because zetafunc had said that earlier
and I'd tested it and seen it fitted the original problem.

So I jumped in with the incorrect 'un-logging' step, and was too easily content
with the result.

I edited that post about 4 times before settling
on the double error (eeeekkkk!) and would have left it
like that if you hadn't posted back with your query.

Thanks for that.

Happy posting!

Bob

Math Is Fun Forum

#1 2011-09-23 20:21:55

Equation problem

#2 2011-09-23 22:25:31

Re: Equation problem

#3 2011-09-23 23:16:07

Re: Equation problem

#4 2011-09-23 23:18:59

Re: Equation problem

#5 2011-09-24 00:46:19

Re: Equation problem

#6 2011-09-24 01:38:54

Re: Equation problem

#7 2011-09-24 04:52:40

Re: Equation problem

#8 2011-09-24 21:25:46

Re: Equation problem

#9 2011-09-24 23:06:59

Re: Equation problem

#10 2011-09-24 23:09:05

Re: Equation problem

#11 2011-09-24 23:53:07

Re: Equation problem

#12 2011-09-24 23:53:56

Re: Equation problem

#13 2011-09-24 23:54:25

Re: Equation problem

#14 2011-09-25 00:02:14

Re: Equation problem

#15 2011-09-25 06:35:24

Re: Equation problem

#16 2011-09-25 07:04:04

Re: Equation problem

#17 2011-09-25 15:50:35

Re: Equation problem

#18 2011-09-28 08:30:28

Re: Equation problem

#19 2011-09-28 09:45:29

Re: Equation problem

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