I've been having problems w/ the equation: "log base x of 3 + log base x of 5 = x." I can use a property of logs to get: "log base x of 15 = x." I can change forms into exponential to get: "x^x = 15." I've tried many different things from here, but nothing seems to definately work. I can use my graphing calculator to get an approximate of x = 2.713... but I wonder if I can somehow get an exact answer because I'm weird. Thanks if you can help.
What rolls down stairs Alone or in pairs Rolls over the neighbore’s dog It’s great for a snack It fits on your back It’s Log, Log, Log! It’s LOG! It’s LOG! It’s big, It’s heavy. It’s wood It’s LOG! It’s LOG! It’s better than bad, it’s good! Log! From Blammo! *Sorry, couldn't help it!*
Yeah, sorry 'bout that.... If I remember correctly, I cheated my way through Trig & Analyt. *did you catch the Ren & Stimpy ref?*
I suck at math, but I have a friend on the phone right now who was a math major in college. I'll channel him for a moment: Okay, he's got it to the x^x=15 stage... . . Now he's swearing. . . He just said that we would have to define a new operation of iterative expontentiation (I don't know how to spell that) and take its inverse. More swearing. He said he'd get back to me. My guess is he'll make an estimate and forget all about it by morning.
I ... actually know what he's talking about ... I wasn't sure if I could use already made up rules, or if I would have to apply effort. You can pretty much do anything you want in math as long as what you do follows basic operations, and has an inverse you can use to get the equation how you want it. In equations like (sqrt x) = 3 .. up until the mid-1700's ... this equation was thought to have no solution because we had no idea anything like a square and sqare root existed [which means that equation would not have even been around, technically] but then we stared to understand the inverse of sqaring and introduced a function that has always been in exsistence, we just haven't always know about it. Same thing here. Like your friend said, you have to "simply" define an inverse for x^x =n to get x = (f^-1(n)) ... That can take some time, abstractity, and calculus sometimes.
Is there definately an analytical solution? I thought about using natural logs, which you can get to: x*ln x = ln 15 or ln x = ( ln 15 ) / x But that still isnt an answer...
Yeah, I thought of that too. It wasn't very helpful. Possibly integrating or taking the derivative? Forms with ln x often simplify that way. I'm not sure it will help in this case though The problem is definitely in terms of log base x, correct?
I bet that's it actually: differentiate it as a product of x and ln x using the product rule. Off the top of my head i think that would be: d/dx(x)*lnx + x*d/dx(lnx) = d/dx(ln15) lnx + x*(1/x) = 0 lnx = -1 x = e^-1 EDIT: I can't even check that because my calculators batteries are dead lol.
hmm ... that's interesting. I try not to use calculus much because I'm still fairly new to it. but I'll try that. By the way, I know e^-1 is incorrect .. or maybe it's an extraneous solution, which means you did nothing wrong. Also, I cooked up an inverse operator [an algorithm that would get me quite close to the answer.] and found out 9 times the log (base 10) of 2 is a VERY close estimate to the real answer.
I had a feeling it might be That said, i can't see any mistakes in my working (although my algebra/calculus is probably a little rusty these days)
I don't know if you took the derivative right, but if I remember calculus at all, solving the derivative of a function for x isn't the same as solving that function for x. x^2 - 4 = 0 gives you x = +/-2. The derivative (I think) would be 2x = 0, and that gives you x = 0.
You're quite right. Having given it some more thought, the reason for that is: We have an expression for x, for which there exists a single solution (maybe multiple solutions) but not a range of solutions. In other words, the expression is not a function. You cannot differentiate under these circumstances, because differentiation calculates rate of change (in this case with respect to x). The expression is constant; it is not changing. Therefore, calculus in this respect, is meaningless. I'll keep thinking
The key is in creating rules to an operation that gives an inverse that gets the x out of the air. Someone from the 1950's created an interesting thing called the "Double Log." It is defined in the case of x^x=n as DBlog (base x) of x = n. [DBlog is pronounced 'double log.'] you can take the log (base t) of each side, and get log (base x-t) of x = log (base t) of n, because you can take the base of the double log and subtract from it the base of the new log you're giving both sides of the equation. For some problems, that is quite useful ... like when you have a word problem and x^x =n has context, but in case the equation in the form of log (base x-t) of x = log (base t) of n just creates confusion, and gets people nowhere. For this equation, I guess I'll have to use iteration to get a good inverse to use.