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If: u = 1+lnx

Then: x = (u-1)/(ln)

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โˆ™ 2011-11-20 15:36:03
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: How do you rearrange u equals 1 plus lnx to get x equals?
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Simplify e to the 1 plus Inx power?

e1 + (lnx) = e1 * e(lnx) = e * x = ex

What is the derivative of y equals xln x?

(xlnx)' = lnx + 1

What is -1 equals lnx?

x = 1/e where e is the base of natural logs.

What is the derivative of lnx raised to lnx?

start by setting y=lnx^lnx take ln of both sides lny=lnx(ln(lnx)) differentiate dy/dx(1/y)=(1+ln(lnx))/x dy/dx=y(1+ln(lnx))/x we know that y=lnx^lnx so we can just substatute back in dy/dx=(lnx^lnx)*(1+ln(lnx))/x

What is the derivative of 1-lnx divided by 2x?


What is the derivative of 1 lnx?

The derivative of 1/lnx, can be found easily using either the chain rule or the quotient rule. It is -1/[x*(lnx)2]

What is the derivative of ln 1 plus x?

d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)

What is the second derivative of x2 ln x?

First one: f'(x) = 2x*lnx + x^2*(1/x) = 2x*lnx + x = x*(2*lnx + 1) Second derivate: f"(x) = D [x*(2*lnx + 1)] = 1*(2*lnx + 1) + x*(2/x) = 2*lnx+1+2 = 2*lnx + 3 So, there is a minimum in this graph on point (1/e^(1/2)), -1/(2e)) = MIN(x, y) Van Sanden David Belgium

What is the derivative of lnx raised to 4?

ln(x4)?d/dx(ln(u))=1/u*d/dx(u)d/dx(ln(x4))=[1/x4]*d/dx(x4)-The derivative of x4 is:d/dx(x4)=4x4-1d/dx(x4)=4x3d/dx(ln(x4))=[1/x4]*(4x3)d/dx(ln(x4))=4x3/x4d/dx(ln(x4))=4/x(lnx)4?Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(lnx)4=4(lnx)4-1*d/dx(lnx)d/dx(lnx)4=4(lnx)3*d/dx(lnx)-The derivative of lnx is:d/dx(ln(u))=1/u*d/dx(u)d/dx(lnx)=1/x*d/dx(x)d/dx(lnx)=1/x*(1)d/dx(lnx)=1/xd/dx(lnx)4=4(lnx)3*(1/x)d/dx(lnx)4=4(lnx)3/x

What is the Derivative of x to the power of x?

The solution to this is: (xx)'= (elnx to the power of x)'= (exlnx)'= (xlnx)'*exlnx= [x(1/x) + 1(lnx)]*exlnx = (lnx+1)*exlnx= (lnx+1)*xx

Does 1 plus 1 plus 1 plus 1 equals 3.8? equals 4...

Derivative of lnx?


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