Solve exactly and make sure to check all solutions: 2LN(x) = LN(2) + LN(3x - 4) x = ?
ln(x^2) = ln(2(3x-4)) Can u do it now?
No I apologize still grasping concepts on logs
Everything else I have done relatively well on until now.
Think...Does it matter what ln is?
Pretend it was 2...
format it in x^2 - 6x - 8 = 0 ?
Yup.
The tricky bit is to get it here ln(x^2) = ln(2(3x-4)) from what u had in your question.
so let me see if I followed what you did
the LN's are basically like terms correct?
so you combined the second two
actually it is x^2 - 6x + 8
so x = 4 or 2
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
aha that helps to get to the first part you solved
and the other....
the first one....
? sorry not understanding what you are asking
I wasn-t asking u anything, I was explaining that I used the first rule as well as the third.
oh alright I understand... thank you
so the final answer is simply 4 and 2?
If that's the solutions to the quadratic, yes.
beautiful... the Natural logs throw me off
you have been a huge help... better then my actual teacher. Been eons since I have been in school
U will need to understand them better when u get further in maths, better do some more studying.
yes been way too long... and have to do some Calc for my CS degree
Definitely need to understand them before u do calc or u will be in trouble
So to make sure I understand. On 7 + 3ln(x) = 6
Ah, this is different...
I wouls change it to 3ln(x) = 6 - 7 so 3ln(x) = -1
Then divide by 3 to get ln(x) = -1/3
Then lost due to ln again lol
No! Use rule 3.
oh then back up
3 ln x = ln x^3
okay yes as you pointed out in rule #3. So x^3 = -1
U could do what you said, this is easier.
No! Again! You only have ln on one side not both.
So you can't just drop it.
or cancel it..
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
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...
By that I mean raise them to the power of e.
which from what a vaguely remember is a ridiculous long number
So u get e^(ln x^3) = e^-1
Now e^(ln x^3) = x^3 (by definition of the natural logarithm).
okay
So you end up with x^3 = e^-1 which u solve for x.
ah wouldn't it be then x = e^ - 1/3 ?
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.
"ah wouldn't it be then x = e^ - 1/3 ?" Yes, this is just a calculation.
To finish, you really do have to get comfortable with this....
you have been incredible
yes that is why I need to keep practicing
Right, do lots of examples till u can do it on autopilot....
actually been doing quite well in class... until these logs
Good luck!
Thank you estudier! hope to see you around again if I need help lol you have been awesome
np anytime.
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