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Solve e to the power ln2?
ln is the inverse of e. So the e and the ln cancel each other out and you are left with 2. eln2 = 2
What is e to the power of 3 ln x?
e^(3lnx)=e^[ln(x^3)]=x^3
What is the derivative of sin x to the e to the xth power?
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
Solve e to the power of 2-x equal 10 to the power of x-2?
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)]
Solve e to the power of 2x plus 2e to power of x equals 8?
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.
Solve e to the power ln2?
ln is the inverse of e. So the e and the ln cancel each other out and you are left with 2. eln2 = 2
What is the derivative of e to the power ln x squared?
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
Simplify e raised to the power of lnx plus lny?
There are 2 interpretations of your question: First: e^[lnx + lny] =e^[ln(xy)] =xy Second: lny + e^(lnx) =lny + x
Simplify ln 1 over 2?
Think of ln 1 as "e to what power will given me 1." Anything to the zero power will give you 1. So, ln 1 = 0, and 0/2 = 0
Simplify In e to the 7th power?
It depends. If you mean (ln e)7, then the answer is 1, since (ln e) = 1. If you mean ln(e7), then the answer is 7, since ln(e7) = 7 (ln e) and (ln e) = 1.
How does e raised to the power of 2x equal e raised to the power of x squared?
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)
Why we cancel e by ln?
The definition of the natural log ln of a number is the power that you have to raise e to in order to get that number. Therefore, ln(2x+3) is the power you have to raise e to to get 2x + 3.
What is e to the power of 3 ln x?
e^(3lnx)=e^[ln(x^3)]=x^3
What is the derivative of x to the power of e?
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
Why 2 power 0 equals 1?
Here is why any number to the zero power equals one. Consider this. a^b. it is natural to restrict a > 0, but we'll only assume that number b is any real number. We'll use the natural exponential function defined by the derivative of the exponential function. Now we have a^r=e^rln(a). And we know that e^rln(a)=e^((ln(a))^r), where a >0 and r is in the domain of all real numbers negative infinity to infinity. We can apply this definition to any number a to any power r. Particularly, a^0. By the provided definition, a^0=e^(0*ln(a))=e^0=1. Furthermore, a^1=e^(1*ln(a))=e(ln(a))=a. And a^2=(e^(ln(a))^2)=a^2.
How do you simplify e raised to 8 ln x plus cos x?
By using the basic rules of exponents, plus the fact that the exponential function (e raised to some power) and the natural logarithm are inverse functions. e8 ln x + cos x = e8 ln x ecosx = e(ln x)(8) ecosx = (eln x)8 ecosx = (eln x)8 ecosx = x8 ecosx
What is the exact solution of e to the x power subtract 2 equals 8?
e^x - 2 = 8 e^x = 10 So x = ln(10) = 1/log10(e)
