1s1 said:Homework Statement
Is ##e^{-ln(x)}## equal to ##\frac{1}{x}## ?
Homework Equations
The Attempt at a Solution
##e^{-ln(x)} = e^{ln(x^{-1})} = e^{ln(\frac{1}{x})} = \frac{1}{x}## ?
Thanks!
e^(-ln(x)) is an expression that represents the inverse function of the natural logarithm. It is equal to 1/x, where x is any positive real number.
To simplify e^(-ln(x)), you can use the property of logarithms that states that ln(e^x) = x. This means that e^(-ln(x)) can be rewritten as e^(ln(1/x)), which simplifies to 1/x.
The significance of e^(-ln(x)) lies in its relationship to the natural logarithm. Since the inverse of the natural logarithm is 1/x, e^(-ln(x)) provides a way to find the multiplicative inverse of any positive real number x.
Yes, e^(-ln(x)) can be used to solve equations involving logarithms. It can be used to rewrite logarithmic equations in exponential form, making it easier to solve for the variable.
Yes, e^(-ln(x)) is always equal to 1/x. This is because the natural logarithm and the exponential function are inverse functions of each other, and the inverse of an inverse is the original function.