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Why we cancel e by ln?

Updated: 4/28/2022
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10y ago

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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.

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Q: Why we cancel e by ln?
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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


If you have a number with the exponent x how do you find the answer?

Take the natural logarithm (ln) of both sides of the equation to cancel the exponent (e). For example, ify=Aexlog transform both sides and apply the rules of logarithms:ln(y)=ln(Aex)ln(y)=ln(A)+ln(ex)ln(y)=ln(A)+xrearrange in terms of x:x=ln(y)-ln(A), or more simplyx=ln(y/A)


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.


What is the value of loge if base is not equal to e?

You can calculate log to any base by using: logb(x) = ln(x) / ln(b) [ln is natural log], so if you have logb(e) = ln(e) / ln(b) = 1 / ln(b)


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


What is e to the power of 3 ln x?

e^(3lnx)=e^[ln(x^3)]=x^3


If e to the power x equals 0.4634 find x?

the natural log, ln, is the inverse of the exponential. so you can take the natural log of both sides of the equation and you get... ln(e^(x))=ln(.4634) ln(e^(x))=x because ln and e are inverses so we are left with x = ln(.4634) x = -0.769165


What is the derivative of e the the power ln x?

y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1


Solve e raised to the power of negative x equals 6?

e-x = 6Take the natural log of both sides:ln(e-x) = ln(6)-x = ln(6)x = -ln(6)So x = -ln(6), which is about -1.792.


How do you do exponential functions?

The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)


How do you differentiate exponential function?

The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)


What is the antiderivative of e to the power of one divided by x?

1/ln(x)*e^(1/x) if you differentiate e^(1/x), you will get ln(x)*e^(1/x). times this by 1/ln(x) and you get you original equation. Peace