solve ln(x-6)+54=ln x-500

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jimmypop

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Apr 26, 2009
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I ran into another logarithmic problem.

I'm not sure if I'm close or not? This is what I was thinking...

ln(x-6)+54=ln x-500
ln(x-6)+554=ln x
ln(x+6/554)=ln x
?? Just not sure if I'm even on the right track?


Sorry if anyones thinking me a pest tonight. I have a test tomorrow and I'm trying to cram all this in. I always get confused with logs.
 
jimmypop said:
ln(x-6)+54=ln x-500
ln(x-6)+554=ln x
ln(x+6/554)=ln x <<< Just derailed! :p

We cannot divide the input to the logarithmic function (i.e., it's argument x + 6). Division by 554 would look like this:

ln(x + 6)/554

In other words, it's the logarithm that's being divided, not the logarithm's argument.
Click to expand...


The strategy on this one is to get both of the logarithms together on one side, and the constant on the other.

We need to combine the logarithmic expression into a single logarithm, in order to then switch the equation's form to exponential notation.

Do you remember the property to write the sum of two logarithms as a single logarithm?

PS: I don't think you're a pest, as long as you're willing to wait for responses (I'm in and out of the backyard, right now).

 
ok, so i have
554 = ln x-500 -ln(x-6)
554 = ln(x-500/x-6)
Can't seem to understand how I'm going to get x by itself?



(I appreciate the help, I'm patient, I still have 7 1/2 hours till class :D and I've already had plenty of sleep today)
 
jimmypop said:
554 = ln(x-500/x-6)
Click to expand...


How did the 500 get back into the righthand side of your previous result?

Let's be clear on the original equation, since your notation is a little inconsistent, and I might have made a bad assumption about it.

I read the original equation as follows, even though you didn't use parentheses to clearly show the logarithm's input x.

ln(x - 6) + 54 = ln(x) - 500

You added 500 to both sides. We can add 500 to both sides only if this is the right interpretation.

ln(x - 500) is a different value, entirely; so, I might not be working with the correct equation. :?

 
Yea the original equation is right, I don;t know what I was thinking?

Let me try that again....

554 = ln(x) - ln(x+6)
554 = ln(x)/ln(x-6)
if this is correct, then I am confused, because wouldn;t the ln(x)'s cancel out? Leaving 554 = 6?


btw, I have to assume (x) is the input for ln as well, The study guide has it written exactly like this.....
ln(x-6)+54=ln x-500
 
jimmypop said:
554 = ln(x) - ln(x+6)

554 = ln(x)/ln(x-6)
Click to expand...


Yeah, this is a common goof. I'll show the property symbolically using two inputs A and B; then I'll correct your version.

ln(A) - ln(B) = ln(A/B)

554 = ln(x/[x - 6])

 
ok since ln(x/(x-6)) Then I can't go any further, because x-6 is in parenthesis?

I think I'm getting the hang of this?
 


Yes, you cannot simplify the lefthand side any more. Now, it's time to switch to exponential form.

ln(x/[x - 6]) = 554

Logarithms are exponents, right? So, when I look at this equation, it's telling me that this logarithm is 554, and, therefore, I realize that it means some exponent is 554.

Can you use the definition of the natural logarithm to rewrite the relationship as an equation that contains a power of e, instead of a logarithm?

(Something seems odd, to me. I think it's time that I actually work this exercise.)

 
??? I am not sure I see how to do that in this problem? I mean I understand that something like log(base e) x=2 can be written x=e^2 but I don't see it here?
 


Excuse the delay; I'm trying to do three things at once.

It seemed odd that we would be working with such a large power of e. Namely, e^554.

I had to use some analysis to determine that neither of the following equations has a solution.

ln(x - 6) + 54 = ln(x) - 500

ln(x - 6) + 54 = ln(x - 500)

I would scrap this exercise; I think the study guide might have a misprint. I'll spend a few minutes thinking about what it could be.

 


I examined the rate of growth for both ln(x - 6) and ln(x), and I can't figure out what they're doing.

How about I make a new equation for you?

ln(x - 6) + 54 = ln(x) + 50

 
jimmypop said:
??? I am not sure I see how to do that in this problem? I mean I understand that something like log(base e) x=2 can be written x=e^2 but I don't see it here?
Click to expand...


Anytime you have a logarithm set equal to something, you can express the relationship between the two using exponential notation.

We have ln(x/[x - 6]) set equal to some constant, C.

ln(x/[x - 6]) = C

Now, make a substitution of symbols.

Let P = x/(x - 6)

ln(P) = C

If you understand that this last equality can be expressed as e^C = P, then you should be able to get the following.

e^C = x/(x - 6)

 


I need to shut down.

Here's the steps for working out the replacement exercise that I posted above.

\(\displaystyle ln(x - 6) + 54 = ln(x) + 50\)

\(\displaystyle ln(x) - ln(x - 6) = 54 - 50\)

\(\displaystyle ln(\frac{x}{x - 6}) = 4\)

\(\displaystyle e^4 = \frac{x}{x - 6}}\)

At this point, I'm making a substitution in symbols (to make the typing easier to read)

Let A = e^4

\(\displaystyle A = \frac{x}{x - 6}}\)

\(\displaystyle A(x - 6) = x\)

\(\displaystyle Ax - 6A = x\)

\(\displaystyle Ax - x = 6A\)

\(\displaystyle x(A - 1) = 6A\)

\(\displaystyle x = \frac{6A}{A - 1}\)

Now, replace A with the value that we have for it: e^4.

\(\displaystyle x = \frac{6e^4}{e^4 - 1}\)

\(\displaystyle x = 6.314374182\)

 
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