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asherwood3
Group Title
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???
 one year ago
 one year ago
asherwood3 Group Title
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???
 one year ago
 one year ago

This Question is Open

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
Divide by 3 on both sides
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
thn carry on
 one year ago

asherwood3 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
Yes.. I will assist you... now what will be the equation?
 one year ago

asherwood3 Group TitleBest ResponseYou've already chosen the best response.0
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..
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
mmm.. you mean I have to write each steps??
 one year ago

asherwood3 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
Ok.. afet dividing by 3 you have to take inverse logarithm
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
then you will be abl to find the value of x
 one year ago

asherwood3 Group TitleBest ResponseYou've already chosen the best response.0
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... :/
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
actually writting step by step or direct answr is against the CoC... So I will write an example
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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". \[\huge x=e^2\]
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
well examole by @zepdrix is sufficient , I guess... You have to work it out
 one year ago

asherwood3 Group TitleBest ResponseYou've already chosen the best response.0
Ohhh ok, so natural Log (Ln) and (e) cancel out. ok. Well that does help out a little bit. Thank you for your time.
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
welcome......
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

asherwood3 Group TitleBest ResponseYou've already chosen the best response.0
Thank you, I'm sure i'll figure it out. Your explanation did help. Yes, Inverse operation of Logarithm is tricky.
 one year ago

Rosh007 Group TitleBest ResponseYou've already chosen the best response.1
all the best....
 one year ago
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