Alright guys I'm really stuck on this one.. I'm trying to solve a Log problem but I cant figure out what step is next...
3Ln(X + 4) -5 = 3
First step: add five to both sides
3Ln (X + 4) -5 = 3
+5 +5
3Ln (x + 4) = 8
What do I do next???

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- anonymous

- jamiebookeater

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- anonymous

Divide by 3 on both sides

- anonymous

thn carry on

- anonymous

so it should look something like this then, right...
3 Ln ( X + 4) = 8
---- ---
3 3
Can you walk me through each step, please? I would really appreciate it

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- anonymous

Yes.. I will assist you... now what will be the equation?

- anonymous

My question is to walk me through the next couple of steps after I divide by 3 on both sides. Can you show me what it should look like? Thanks..

- anonymous

mmm.. you mean I have to write each steps??

- anonymous

well if you could just briefly explain to me what to do after I divide by 3 that would be helpful.... I'm just completely lost on this problem. I pretty much only know how to do the first step

- anonymous

Ok.. afet dividing by 3 you have to take inverse logarithm

- anonymous

then you will be abl to find the value of x

- anonymous

so what might that look like... could you write out the steps by placing the numbers in the correct spots. I know I'm asking a lot but it would really help me understand the problem better... :/

- anonymous

actually writting step by step or direct answr is against the CoC...
So I will write an example

- zepdrix

Exponentiation is the inverse of the logarithmic function. Here's a quick example.
\[\large \ln x = 2\]
This is a log with base e, we'll exponentiate both sides, rewrite both sides as exponents with base e.
\[\huge e^{\ln x}=e^2\]
Since the exponential and the logarithm are inverse operations of one another, they essentially "cancel out".
\[\huge x=e^2\]

- anonymous

well examole by @zepdrix is sufficient , I guess... You have to work it out

- anonymous

Ohhh ok, so natural Log (Ln) and (e) cancel out. ok. Well that does help out a little bit. Thank you for your time.

- anonymous

welcome......

- zepdrix

Yah performing the inverse operation of the logarithm is a bit tricky to get used to :)
If you're still confused on how to perform that on your particular problem, just let us know.

- anonymous

Thank you, I'm sure i'll figure it out. Your explanation did help. Yes, Inverse operation of Logarithm is tricky.

- anonymous

all the best....

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