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ln is the natural logarithm. That is it is defined as log base e. As we all know from school, log base 10 of 10 = 1 just as log base 3 of 3 = 1, so, likewise, log base e of e = 1 and 1.x = x.
so we have ln y = x. Relace ln with log base e, and you should get y = ex
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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 Differentiate and Integrate y equals 3 to the power of x?
dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).
How to solve e to the power of 3x plus 5e to the power of x equals 7?
e3x+5 x ex =7 e3x+5+x=7 4x+5=ln(7) x=(ln(7)-5)/4
What is the answer to p equals 4 - ln x?
p=4-lnx Are you trying to solve for x? If so, then lnx=4-p so x=e^(4-p) Perhaps you left out some information?
What is the natural log of of 4x equals 13?
ln(4x)=13 4x=e^13 x=(e^13)/4
How do you solve 8958 equals e5x?
8958=e^(5x) ln both sides -> ln(8958)=5x Therefore x=1.82
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 Differentiate and Integrate y equals 3 to the power of x?
dy/dx = 3^x * ln(3)integral = (3^x) / ln(3)To obtain the above integral...Let y = 3^xln y = x ln 3y = e^(x ln 3)(i.e. 3^x is the same as e^(x ln 3) ).The integral will then be 3^x / ln 3 (from linear composite rule and substitution after integration).
How do you solve this logarithmic equation ln x equals 5?
In the equation ln(x) = 5, the solution is x = (about) 148.4. To solve, simply raise e to the power of both sides and reduce... ln(x) = 5 eln(x) = e5 x = 148.4
How do you solve 3e to the power 2x -1 equals 8?
you need to know natural logarithms3e to the 2x-1 power = 8(2x-1) ln e = ln (8/3)ln e = 1(2x-1) = ln(8/3) = 0.982x = 1.98x = 0.99
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 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
How to solve e to the power of 3x plus 5e to the power of x equals 7?
e3x+5 x ex =7 e3x+5+x=7 4x+5=ln(7) x=(ln(7)-5)/4
How do you solve 3 ln x - ln 2 equals 4?
3lnx - ln2=4 lnx^3 - ln2=4 ln(x^3/2)=4 (x^3)/2=e^4 x^3=2e^4 x=[2e^4]^(1/3)
What is the answer to p equals 4 - ln x?
p=4-lnx Are you trying to solve for x? If so, then lnx=4-p so x=e^(4-p) Perhaps you left out some information?
How do you solve 5 to the power of x equals 15?
To solve the equation 5^x = 15, you can take the logarithm of both sides. By taking the natural logarithm of both sides, you get x * ln(5) = ln(15). Then, you can solve for x by dividing both sides by ln(5), giving you x = ln(15) / ln(5), which is approximately 1.682.
Is there a function where the first derivative goes to infinity for x going to 0 and where the first derivative equals 0 when x is 1?
Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.
