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Q: How would you solve ln 4 plus 3 ln x equals 5 ln 2?

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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)

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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

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.

if x8 = 20 then x = - eighth root of 20 or x = eighth root of 20 eighth root of 20 = 1.4542154.... Maybe you mean 8x = 20 If so Ln (8x) = Ln (20) => x Ln(8)= Ln(20) => x = Ln(20)/Ln(8) = 1.4406426...

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5 + (-4) = 1 x = -4

-3 + ln(x) = 5 ln(x) = 8 eln(x) = e8 x = e8 x =~ 2981

2 ln(9) + 2 ln(5) = 2 ln(x) - 3ln(81) + ln(25) = ln(x2) - 37.61332 = ln(x2) - 3ln(x2) = 10.61332ln(x) = 5.30666x = e5.30666 = 201.676 (rounded)

3 ln(x) = ln(3x)ln(x3) = ln(3x)x3 = 3xx2 = 3x = sqrt(3)x = 1.732 (rounded)

8958=e^(5x) ln both sides -> ln(8958)=5x Therefore x=1.82

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There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).

e3x+5 x ex =7 e3x+5+x=7 4x+5=ln(7) x=(ln(7)-5)/4

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

If: u = 1+lnx Then: x = (u-1)/(ln)

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 solve ln|tan(x)|=ln|sin(x)|-ln|cos(x)|? Well you start by........

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