Let f=exp(x) and g=ln(x)Maybe identities isn't the right word,...
JJacquelin said:Let f=exp(x) and g=ln(x)
Of course there are basic relationships :
x = ln(f) = exp(g)
ln(exp(x)) = exp(ln(x))
I'm looking for a way to substitute [tex]{1/ln(x)}[/tex] with an exponential function to help solve an integral (the integral, I've asked about it here before, is [tex]\int sin(x)/ln(x)\,dx[/tex] sorry, don't know how to do fractions within integrals in latex!). Only substitution I can think of is [tex]{1/ln(x)=\log_x e}[/tex]. However, that won't help since it's still a logarithmic function!micromass said:Well, since these inverses are the definition of the logarithm, it follows that any other identity can be derived from those. So I'm not sure what kind of identities you want?? It would help us if you told us what you are looking for and why...
romsofia said:I'm looking for a way to substitute [tex]{1/ln(x)}[/tex] with an exponential function to help solve an integral (the integral, I've asked about it here before, is [tex]\int sin(x)/ln(x)\,dx[/tex] sorry, don't know how to do fractions within integrals in latex!). Only substitution I can think of is [tex]{1/ln(x)=\log_x e}[/tex]. However, that won't help since it's still a logarithmic function!
Thanks for showing on how to do fractions in integralsmicromass said:The integral (right click to see how I did the fraction)
[tex]\int{\frac{\sin(x)}{ln(x)}dx}[/tex]
is not solvable using elementary functions, so you won't be able to solve it. The best thing to do is to find a series solution, which is also not easy.
romsofia said:Thanks for showing on how to do fractions in integrals
Well, I was thinking if we were to go to the complex plane for the problem, then maybe we could solve it? Since we can make the substitution [tex]{sin(x)=1/2i(e^{-ix}-e^{ix})}[/tex] However, I haven't had much exposure to complex numbers so I'm not really sure if it would make a difference if we were to work in the complex plane! However, I'm still lost on any exponential functions that we would be able to substitute for [tex]{1/ln(x)}[/tex]
micromass said:Going to the complex numbers wouldn't make any difference. Mathematica uses an algorithm that decides 100% if there is a solution, and if it says that there isn't, then there isn't (even when working with complex numbers).
The relationship between exponential and logarithmic functions is that they are inverse functions of each other. This means that they "undo" each other's operations, and can be used to solve for the unknown variable in an equation.
The main difference between exponential and logarithmic functions is the way in which the variable is used. In exponential functions, the variable is in the exponent, while in logarithmic functions, the variable is the base of the logarithm.
To graph an exponential function, you can use the general form y = ab^x, where a is the initial value and b is the growth factor. For logarithmic functions, the general form is y = logb(x), where b is the base of the logarithm. You can plot points and connect them to create a curve for both types of functions.
Exponential functions can be used to model population growth, compound interest, and radioactive decay. Logarithmic functions can be used to measure the intensity of earthquakes, the pH level of a substance, and the loudness of sound.
To solve equations with both exponential and logarithmic functions, you can use the property that they are inverse functions. This means that you can rewrite the equation using logarithms or exponents to "undo" the other function. Then, you can solve for the variable using algebraic techniques.