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asherwood3
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???
Divide by 3 on both sides
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
Yes.. I will assist you... now what will be the equation?
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..
mmm.. you mean I have to write each steps??
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
Ok.. afet dividing by 3 you have to take inverse logarithm
then you will be abl to find the value of x
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... :/
actually writting step by step or direct answr is against the CoC... So I will write an example
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". \[\huge x=e^2\]
well examole by @zepdrix is sufficient , I guess... You have to work it out
Ohhh ok, so natural Log (Ln) and (e) cancel out. ok. Well that does help out a little bit. Thank you for your time.
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.
Thank you, I'm sure i'll figure it out. Your explanation did help. Yes, Inverse operation of Logarithm is tricky.