NO! Lnx + Ln2= 2 + Lnx implies Ln2 = 2 which implies 2 = e2 which is simply not true.
lnx + lnx +3 = (ln55)/4 2lnx +3 =(ln55)/4 8lnx + 12=ln55 8lnx=-12+ln55 lnx=(-12+ln55)/8 x=e^[(-12+ln55)/8]
i believe it is 7lnx, but don't quote me on it.
If: u = 1+lnx Then: x = (u-1)/(ln)
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e1 + (lnx) = e1 * e(lnx) = e * x = ex
NO! Lnx + Ln2= 2 + Lnx implies Ln2 = 2 which implies 2 = e2 which is simply not true.
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
lnx + lnx +3 = (ln55)/4 2lnx +3 =(ln55)/4 8lnx + 12=ln55 8lnx=-12+ln55 lnx=(-12+ln55)/8 x=e^[(-12+ln55)/8]
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
i believe it is 7lnx, but don't quote me on it.
If: u = 1+lnx Then: x = (u-1)/(ln)
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The answer is "no solution." ln(0) means e^x=0, and nothing raised to any power can ever equal zero. The domain of y=lnx is (0,∞) and the range is (-∞,∞).
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
I get x*x^x-1 + lnx*x^x = x^x + x^xlnx = x^x * (1+lnx) Here, ^ is power; * = times; ln = natural logratithm ( base e)
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