how do i simplify e^(1 + lnx)?

Answer 1

e^(1 + lnx)

we know that ln(e) = 1

e^(ln(e) + ln(x) )

product rule: ln a + ln b = ln (ab)

e^(ln(e*x))

e^(ln x) = x

so the answer is e*x

Answer 2

An alternate method is to realize that adding exponents is just combining a series of values which have been multiplied, e.g.

2^2 * 2^3 = 2(2+3) = 2^5 = 32

So, you can work in reverse for your question:

e^(1 + lnx) = e^1 * e^lnx = e*x (since with e^lnx the e and ln cancel to leave x)

Answer 3

e^(1 + lnx) = e^1 · e^(lnx) = e*x

Saludos.

Answer 4

e^ needed -lnx would desire to be written as e^(needed -ln(x) dx) =e^(-needed ln(x) dx) Now, by using factors, needed (lnx)dx = xln(x)-x+C_1, the situation C_1 is the consistent of integration. So e^(-needed ln(x) dx) = e^(-xln(x)-x+C_1)=e^(-xln(x))*e^(-x)*e^C... =C_2x^(-x)*e^(-x), the situation C_2=e^C_1 is a non-0 consistent, and e^(-xln(x))=x^(-x). suited variety is C_2(xe)^(-x).

Answer 5

e^(1+ln x) = e*e^(ln x) = e*x.