Solving Logs: Step-by-Step Guide

[elemis]

Is there a general method for solving these kind of questions ?

How would you solve it ? With steps please.

log x= 5^-x

log x=3^x-3

 

[elemis]

Anyone ?

 

[elemis]

Come now. Dont be shy :P

 

[Theaumasch]

You need to solve them numerically. Topics like this one belong in the homework forums.

 

[elemis]

You need to solve them numerically. Topics like this one belong in the homework forums.

Please do explain further. At least solve and show me the steps.

 

[Theaumasch]

1. Get your graphing calculator.
2. Y1=5^-x, Y2=log(x)
3. Plot and use the intersect function of your calculator.

 

[Mentallic]

They can be solved using the Lambert W function, but if you haven't heard of ths before then it won't really be of any help since that function is solved numerically anyway.

 

[mathwonk]

you could graph the two equations and try to see where they cross. Or you could try to find a zero of their difference by Newton's method. one trivial comment is that x > 0, since other wise log(x) is undefined. In fact since 5^(-x) > 0, and log(x) is only positive when x > 1, we know x > 1. in fact when x = 2, we have 5^(-x) = 1/25 = .04, and since log(2) is probably bigger than that (isn't it about .6? Anyway just looking at ln(1+x) = the integral of 1/(1+x) shows it is easily more than .5) then 1 < x < 2. so just get in there and start trying elementary estimates to get a feel for it. (I presume log means ln).

 

[mathwonk]

Basic rule, only trivial "cook book" equations can be "solved" explicitly, so Newton's method is always the way to go in general.

 

[HallsofIvy]

They can be solved using the Lambert W function, but if you haven't heard of ths before then it won't really be of any help since that function is solved numerically anyway.

Just like log, or exp, or sine and cosine.

 

[Mentallic]

Just like log, or exp, or sine and cosine.

Yes but at least most ordinary hand calculators have the other functions. I can't find my W function on mine

 

[jackmell]

They can be solved using the Lambert W function, but if you haven't heard of ths before then it won't really be of any help since that function is solved numerically anyway.

I do not believe those two equations can be solved for x in terms of the Lambert-W function because the inverse functions work against any sort of isolation of x.

 

[Mentallic]

I do not believe those two equations can be solved for x in terms of the Lambert-W function because the inverse functions work against any sort of isolation of x.

Oh yes you're right! How silly of me

 

[jackmell]

Oh yes you're right! How silly of me

Hi Mentallic. To be fair, I first tried to solve them in Mathematica. Now, just because it couldn't do it is no guarantee but that did give me some confidence the W function could probably not be used here. I recommend to everyone here to try and get use to using Mathematica to confirm your work but not rely on it to much. :)

 

[Mentallic]

Mathematica is pretty damn good, so if you find that it doesn't give you an exact solution in terms of the function, it should make you start thinking that way too.

The W function accounts for equations of the form $$x=a^x$$ but not of $$x=a^{b^x}$$ which is what you would get from taking the exponential of both sides. I quickly skimmed over the questions and the W functions quickly came to mind. I should have taken a closer look at them though.

 

[uart]

The Lambert W function is very useful for solving both log-linear and exp-linear equations such as :

$$ln(x) + ax = b$$

or

$$e^x + ax = b$$

For the log type the solution is :

$$x = \frac{1}{a} \, W\left(a e^b\right)$$

And for the exp type the solution is :

$$x = \frac{b}{a} - W\left(\frac{1}{a} \, e^{b/a}\right)$$

For more complicated log and exponential equations I find that sometimes I can manipulate them into a form that's amenable to the Lambert W function, but often not.

 

[uart]

Re getting the Lambert W function on a standard pocket calculator.

I've found that Lambert W is pretty amenable to simple "fix point iteration", so if you can solve it fairly easily on most (non programmable) pocket calculators.

As you know the value of $W(c)$, for some constant "c", is the solution to $x e^x = c$. To solve by fixed point iteration simply rearrange it into the form of x = f(x) in either of the most obvious ways. That is either of

1. $$\,\,\,\,\,x = \frac{c}{e^x}$$

or

2. $$\,\,\,\,\,x = ln(\frac{c}{x})$$

Rearrangement #1 converges under fixed point iteration for $c<e$ and rearrangement #2 converges for $c>e$.

Unfortunately the convergence of both is poor for arguments around e but with a simple "averaging" modification the second one does converge quote nicely around e. That is,

3. $$\,\,\,\,\,x = \frac{1}{2} \left( x + ln(\frac{c}{x}) \right)$$

Rearrangement number #3 is best used for arguments "c" in the range of about $1 < c < 10$.