- #1
JungleJesus
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Just out of curiosity, I was wondering how to solve for x in an expression of the form
ax = xb
It kinda tripped me out at first.
ax = xb
It kinda tripped me out at first.
Yes. Taking the "b"th root of both sides, [itex](a^{1/b})^x= x[/itex] so that [itex]1= xe^{-(1/b)(ln a)x}[/itex]. Multiply on both sides by -(1/b)ln a: [itex]-(1/b)ln a[/itex][itex]= (-(1/b)(ln a)x)e^{-(1/b)(ln a)x}[/itex].Anonymous217 said:You can try manipulating it to fit the lambert W function I think.
HallsofIvy said:Letting y= -(1/b)(ln a)x, that becomes [itex]ye^y= -(1/b)ln a[/itex]. Using the Lambert W function (which is defined as the inverse function to f(x)= xex) we have y= W(-(1/b)ln a).
Then [itex]x= -\frac{b}{ln a}W(-(1/b)ln a)[/itex]
JungleJesus said:Is there any way to simplify this expression, or is this the bare bones? How would one procede to calculate this value?
The first step is to rewrite the equation so that x is only either an exponent or a base, but not both. For example, if you have an equation like x^x = 4, you can rewrite it as x = 4^(1/x). Then, you can solve for x using algebraic manipulation or trial and error.
Yes, you can use logarithms to solve for x in equations where x is both an exponent and a base. You can use the logarithm rule log(a^b) = b*log(a) to rewrite the equation and then solve for x using the properties of logarithms.
Yes, there are other methods such as substitution, factoring, or using the quadratic formula. The best method to use will depend on the specific equation and your comfort level with each method.
Yes, there are a few special cases to consider. One is when the base is a negative number, in which case the solution may involve imaginary numbers. Another is when the exponent is a fraction, in which case the solution may involve taking the root of a number. It is important to check for these special cases and handle them accordingly when solving for x.
To check if your solution is correct, you can substitute your value for x back into the original equation and see if it satisfies the equation. Another way is to graph the equation and your solution on a coordinate plane to see if they intersect at the same point.