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Mathematics 21 Online
OpenStudy (anonymous):

Solve exactly and make sure to check all solutions: 2LN(x) = LN(2) + LN(3x - 4) x = ?

OpenStudy (anonymous):

ln(x^2) = ln(2(3x-4)) Can u do it now?

OpenStudy (anonymous):

No I apologize still grasping concepts on logs

OpenStudy (anonymous):

Everything else I have done relatively well on until now.

OpenStudy (anonymous):

Think...Does it matter what ln is?

OpenStudy (anonymous):

Pretend it was 2...

OpenStudy (anonymous):

format it in x^2 - 6x - 8 = 0 ?

OpenStudy (anonymous):

Yup.

OpenStudy (anonymous):

The tricky bit is to get it here ln(x^2) = ln(2(3x-4)) from what u had in your question.

OpenStudy (anonymous):

so let me see if I followed what you did

OpenStudy (anonymous):

the LN's are basically like terms correct?

OpenStudy (anonymous):

so you combined the second two

OpenStudy (anonymous):

actually it is x^2 - 6x + 8

OpenStudy (anonymous):

so x = 4 or 2

OpenStudy (anonymous):

Here are some of the rules for ln's (natural logarithms)(or logs in general). log(xy) = log x + log y log(x/y) = logx - log y log x^p = p log x

OpenStudy (anonymous):

aha that helps to get to the first part you solved

OpenStudy (anonymous):

and the other....

OpenStudy (anonymous):

the first one....

OpenStudy (anonymous):

? sorry not understanding what you are asking

OpenStudy (anonymous):

I wasn-t asking u anything, I was explaining that I used the first rule as well as the third.

OpenStudy (anonymous):

oh alright I understand... thank you

OpenStudy (anonymous):

so the final answer is simply 4 and 2?

OpenStudy (anonymous):

If that's the solutions to the quadratic, yes.

OpenStudy (anonymous):

beautiful... the Natural logs throw me off

OpenStudy (anonymous):

you have been a huge help... better then my actual teacher. Been eons since I have been in school

OpenStudy (anonymous):

U will need to understand them better when u get further in maths, better do some more studying.

OpenStudy (anonymous):

yes been way too long... and have to do some Calc for my CS degree

OpenStudy (anonymous):

Definitely need to understand them before u do calc or u will be in trouble

OpenStudy (anonymous):

So to make sure I understand. On 7 + 3ln(x) = 6

OpenStudy (anonymous):

Ah, this is different...

OpenStudy (anonymous):

I wouls change it to 3ln(x) = 6 - 7 so 3ln(x) = -1

OpenStudy (anonymous):

Then divide by 3 to get ln(x) = -1/3

OpenStudy (anonymous):

Then lost due to ln again lol

OpenStudy (anonymous):

No! Use rule 3.

OpenStudy (anonymous):

oh then back up

OpenStudy (anonymous):

3 ln x = ln x^3

OpenStudy (anonymous):

okay yes as you pointed out in rule #3. So x^3 = -1

OpenStudy (anonymous):

U could do what you said, this is easier.

OpenStudy (anonymous):

No! Again! You only have ln on one side not both.

OpenStudy (anonymous):

So you can't just drop it.

OpenStudy (anonymous):

or cancel it..

OpenStudy (anonymous):

if you could hear how angry at myself I am right now you would appreciate it. Again been too long so your patience is incredible

OpenStudy (anonymous):

This is why you need to understand what a log (and especially a natural log) is. U have ln x^3 = -1 and now u have to "exponentiate" both sides...

OpenStudy (anonymous):

By that I mean raise them to the power of e.

OpenStudy (anonymous):

which from what a vaguely remember is a ridiculous long number

OpenStudy (anonymous):

So u get e^(ln x^3) = e^-1

OpenStudy (anonymous):

Now e^(ln x^3) = x^3 (by definition of the natural logarithm).

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

So you end up with x^3 = e^-1 which u solve for x.

OpenStudy (anonymous):

ah wouldn't it be then x = e^ - 1/3 ?

OpenStudy (anonymous):

The key thing u need to understand is that x = log to the base a of y MEANS y = a^x (ie the log is the inverse of the exponential). In the case of natural logs, the base is e.

OpenStudy (anonymous):

"ah wouldn't it be then x = e^ - 1/3 ?" Yes, this is just a calculation.

OpenStudy (anonymous):

To finish, you really do have to get comfortable with this....

OpenStudy (anonymous):

you have been incredible

OpenStudy (anonymous):

yes that is why I need to keep practicing

OpenStudy (anonymous):

Right, do lots of examples till u can do it on autopilot....

OpenStudy (anonymous):

actually been doing quite well in class... until these logs

OpenStudy (anonymous):

Good luck!

OpenStudy (anonymous):

Thank you estudier! hope to see you around again if I need help lol you have been awesome

OpenStudy (anonymous):

np anytime.

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