e2ln2 + 2 = 22 + 2 = 4 + 2 = 6
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ln is the inverse of e. So the e and the ln cancel each other out and you are left with 2. eln2 = 2
e^(3lnx)=e^[ln(x^3)]=x^3
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
e^2-x = 10^x-2 let take the natural log of both sides we have 2-x=ln ( 10^(x-2) 2-x=x-2(ln 10) 4-2x=ln10 -2x=ln10-4 x= -1/2[(ln10)+4)]
e2x + 2ex = 8. First, let's define z to be ex Substituting gives us the following quadratic equation:z2+2z=8 or z2+2z-8=0Solving for z gives us two possible zs: 2 and minus 4.As z is a power of e, only z=2 makes sense.Back solving we get:ex = 2and the answer is x=ln(2). If you substitute in the answer, you get e2ln2+2eln2=8, or 4+4=8. ln(2) ~= 0.69315.