How do i solve an equation like e^(1+lnx)?

Question

Hi guys i am in precalc but i am having trouble solving an equation like e^ (1+lnx)
your earliest response is appreciated...
please help!!!!

Answer 1

You need to know that e^(a+b) = (e^a)*(e^b).

So...
e^(1+lnx) =
e^1 * e^lnx

Then you need to know that e and ln are inverse functions, so e^lnx = x:

e^1 * e^lnx =
e * x

Answer 2

It is not an equation - no equal sign. You can simplify it:
e^(1+lnx)=e^1 * e^lnx = ex

Answer 3

This can be simplified down to:

(e^1)*(e^lnx)

because you add exponents when you multiply. This can be simplified further:

(e^1)*x

because e^lnx = x

Answer 4

break it into a series equivalent.

or break it into its components.

if i remember e^(a+b) = e^a * e^b
which would make this

(e^1) * (e^lnx)
e^1 = e

e^lnx = x (i believe)

which would make this x*e

lemme know
[e-mail]

Answer 5

This is not an equation. This is an expression to be simplified.

e^(1 + ln x)

= (e^1)( e^(ln x)) because a^(m + n) = (a^m)(a^n)

= (e)(x) because a^1 = a and e^(ln a) = a.

Answer: ex or xe (the product of x e and x)

Answer 6

Remember that the function's domain is restricted!!!!! For all x>0, it is the function f(x) = e*x, as the previous posts said. But for x<0, the function becomes imaginary.

Answer 7

review your exponent rules


e^(a+b) = e^a*e^b

so e^(1+lnx) = e^1*e^lnx = e*x

Answer 8

You can use your exponential rules:
e * e^lnx => e * x