find the inverse?

Question

can someone help me to find the inverse of e^x and 2^x ?

Answer 1

y=e^x
LNy=LNe^x=xLNe=x
x=LNy

y=2^x
LNy=LN2^x=xLN2
x=(LNy)/(LN2)

Answer 2

Let f: R -> R be defined as f(x) = e^x, where R is the set of real numbers. Since f(x) is an exponential function, it follows that f has a one-to-one correspondence from its domain R to its codomain R, this can be verified by plotting f(x) in the Cartesian plane.. Thus f has an inverse.

Note that the range of the function f is the set of positive real numbers, which is only the subset of the codomain of f.

To obtain the inverse of f, let y = f(x). [Note that the steps here might differ from the source I cited below, but it will yield to the same result.] We interchange the variables x and y, and then solve for y. That is:
x = f(y) = e^y.

Taking natural logarithm to both sides of the equation gives,
ln x = ln ( e^y )
ln x = y ln ( e )
ln x = y.

which gives the inverse of e^x, i.e. f-1(x) = ln x.

To check whether f and f-1 are inverses of each other, we have
(f o f-1)(x) = f( f-1(x) ) = f ( ln x ) = e^(ln x) = x and
(f-1 o f)(x) = f-1( f(x) ) = f ( e^x ) = ln ( e^x ) = x ln (e) = x.
Thus, f and f-1 are inverses of each other.

Likewise, let g: R -> R be defined as g(x) = 2^x. With the same reasoning as above, we conclude that g is invertible. So to obtain its inverse, let z = g(x). Then we interchange the variables x and z, and then solve for z. That is:
x = g(z) = 2^z.

Taking the logarithm to both sides of the equation gives,
log x = log ( 2 ^ z )
log x = z log 2
z = ( log x ) / ( log 2 ) = lg x,

where lg x is the logarithm of x taken to the base 2. Thus, the inverse of 2^x, is defined by the function g-1(x) = lg x.

To check whether g and g-1 are inverses of each other, we have
(g o g-1)(x) = g( g-1(x) ) = g ( lg x ) = 2^(lg x) = x and
(g-1 o g)(x) = g-1( g(x) ) = g ( 2^x ) = lg ( 2^x ) = x lg (2) = x (log 2) / (log 2) = x.
Thus, g and g-1 are inverses of each other.

Answer 3

inverse of e^x is ln y, i.e. natural log of y
and the inverse of 2^x is log base 2 of y

Answer 4

ln x and ln x/ln 2, respectively.

Answer 5

Let x = e^y
ln(x) = ln(e^y) = y ln(e) = y, so
inv(e^x) = ln(x)

Let x = 2^y
ln(x) =yln(2)
y = (ln(x))/(ln(2))
inv(2^x) = (ln(x))/(ln(2))