this seems wrong but why?

You're forgetting that [MATH]\ln3e^x[/MATH] is not [MATH]\ln(3e)^x[/MATH]. The exponent applies only to [MATH]e[/MATH], not to [MATH]3e[/MATH].

[MATH]\ln(3e)^x[/MATH] simplifies to [MATH]x\ln(3e)[/MATH], because the argument of the log is a single power.

[MATH]\ln3e^x[/MATH] is the log of a product, so it simplifies to [MATH]\ln3 + \ln e^x = \ln3 + x[/MATH].

I have no idea what you did on the third line! Give the whole thing another try.

An important thing to keep in mind is that logs are just numbers, so that something like [MATH]x + \ln 4 = x\ln 3[/MATH] (if that had been correct) can be solved the same way as, say, [MATH]x + 4 = 3x[/MATH], by collecting terms with x on one side. I would hold off on dividing until the last step.
 
allegansveritatem said:
Here is the equation to be solved:
View attachment 15252

Here is what I did:
'View attachment 15253

Where id I go wrong here?
Click to expand...
The equation went wrong.. It doesn't work.

Let me explain it a bit more simply:
[math]e^{x + ln(4)} = 3 e^x[/math]
[math]e^x \cdot e^{ln(4)} = 3 e^x[/math]
Cancelling out the common [math]e^x[/math] gives
[math]e^{ln(4)} = 3[/math]
which is false. There is no solution to this equation. (You can do it your way as well, in the form suggested by Dr.Peterson if you like.)

-Dan
 
allegansveritatem said:
Here is the equation to be solved:
View attachment 15252

Here is what I did:
'View attachment 15253

Where id I go wrong here?
Click to expand...
Let's talk about the error in the 3rd line as it is important to see what went wrong (even though you made an earlier error).

The rhs was \(\displaystyle \dfrac{x+ln(4)}{ln(3e)}\) and you want to divide this by x. Well \(\displaystyle \dfrac{x + ln(4)}{ln(3e)}\) = \(\displaystyle \dfrac{x}{ln(3e)}+ \dfrac{ln(4)}{ln(3e)}\) Now I understand that \(\displaystyle \dfrac{ln(4)}{ln(3e)}\) divided by x is \(\displaystyle \frac{\frac{ln(4)}{ln(3e)}}{x}\) OR \(\displaystyle \dfrac{ln(4)}{xln(3e)}\) but what happened to \(\displaystyle \dfrac{\dfrac{x}{ln(3e)}}{x}?\) That should equal \(\displaystyle \dfrac{1}{ln(3e)}\neq0\)
 
Jomo said:
Let's talk about the error in the 3rd line as it is important to see what went wrong (even though you made an earlier error).

The rhs was \(\displaystyle \dfrac{x+ln(4)}{ln(3e)}\) and you want to divide this by x. Well \(\displaystyle \dfrac{x + ln(4)}{ln(3e)}\) = \(\displaystyle \dfrac{x}{ln(3e)}+ \dfrac{ln(4)}{ln(3e)}\) Now I understand that \(\displaystyle \dfrac{ln(4)}{ln(3e)}\) divided by x is \(\displaystyle \frac{\frac{ln(4)}{ln(3e)}}{x}\) OR \(\displaystyle \dfrac{ln(4)}{xln(3e)}\) but what happened to \(\displaystyle \dfrac{\dfrac{x}{ln(3e)}}{x}?\) That should equal \(\displaystyle \dfrac{1}{ln(3e)}\neq0\)
Click to expand...
I have been looking and looking at that third line and I'm damned if I know what I was thinking about when I did that. I vaguely recall that I elided a step at that point....but now I just don't know what the elision consisted of. Where did I get that one? I am not sure what you mean with the terms at the end of your reply either....I will go over the whole thing again tomorrow and get back. Thanks for checking it out.
 
Dr.Peterson said:
You're forgetting that [MATH]\ln3e^x[/MATH] is not [MATH]\ln(3e)^x[/MATH]. The exponent applies only to [MATH]e[/MATH], not to [MATH]3e[/MATH].

[MATH]\ln(3e)^x[/MATH] simplifies to [MATH]x\ln(3e)[/MATH], because the argument of the log is a single power.

[MATH]\ln3e^x[/MATH] is the log of a product, so it simplifies to [MATH]\ln3 + \ln e^x = \ln3 + x[/MATH].

I have no idea what you did on the third line! Give the whole thing another try.

An important thing to keep in mind is that logs are just numbers, so that something like [MATH]x + \ln 4 = x\ln 3[/MATH] (if that had been correct) can be solved the same way as, say, [MATH]x + 4 = 3x[/MATH], by collecting terms with x on one side. I would hold off on dividing until the last step.
Click to expand...
yes, I see that. Right, the factors can be separated. As for that third line, I have been puzzling over it myself. I am flummoxed by it. What was I thinking about? I am going to work on the thing again tomorrow and see what I can come up with. Thanks for the tip re to what the exponent pertains.
 
topsquark said:
The equation went wrong.. It doesn't work.

Let me explain it a bit more simply:
[math]e^{x + ln(4)} = 3 e^x[/math]
[math]e^x \cdot e^{ln(4)} = 3 e^x[/math]
Cancelling out the common [math]e^x[/math] gives
[math]e^{ln(4)} = 3[/math]
which is false. There is no solution to this equation. (You can do it your way as well, in the form suggested by Dr.Peterson if you like.)

-Dan
Click to expand...
Well, I see you are right. If I had separated the components on the left side as you did, I would have seen that. Ln 4 is certainly not equal to three. It must be that the author of this problem threw it in as one might throw a wrench into the works---just to liven things up a bit. Thanks for pointing this out. I will still work it out for myself tomorrow, but I can see that the jig is up for it already..
 
I went at it again today, even though by then I know that the thing was insoluble but I wanted to prove it to myself and I did it in the way topsquark did it above and then I did along the lines that I was following when I thought it could be solved--just to see what I was thinking about when came up with those strange indecipherable terms that were in the third line of my original post. It seems that I got the 1 in the third line of that post by dividing x by x!
esolution1.PNG

esolution2.PNG
 
First of all getting ln(3e)/ln(4) = 1 is a good thing as you were to get a contradiction. After topsquark got e^(ln(4)) = 3.
The problem is that you made mistakes at arriving at ln(3e)/ln(4) = 1

Since when do you undo addition of ln(4) by dividing by ln(4)? Also Dr Peterson told you that you ln(3e^x)\(\displaystyle \neq\)xln(3e)
 
Jomo said:
First of all getting ln(3e)/ln(4) = 1 is a good thing as you were to get a contradiction. After topsquark got e^(ln(4)) = 3.
The problem is that you made mistakes at arriving at ln(3e)/ln(4) = 1

Since when do you undo addition of ln(4) by dividing by ln(4)? Also Dr Peterson told you that you ln(3e^x)\(\displaystyle \neq\)xln(3e)
Click to expand...
Yes, I see what you are saying. I again made the mistake of not separating the factors 3 and e^x.
 
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