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)
ln2^x = xln2. let ln2 = k (constant), then the differential = k. Hence d(ln2^x)/dx = ln2
E is energy and MC2 is matter used to the second power... now you solve it
PLUS 2
E to the power infinity, or lim en as n approaches infinity is infinity.
In (x^(ln2)), ln2 is a constant, so the Power Rule can be used. d/dx (x^ln2)=(ln2)x^(ln2-1)
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)
ln2^x = xln2. let ln2 = k (constant), then the differential = k. Hence d(ln2^x)/dx = ln2
NO! Lnx + Ln2= 2 + Lnx implies Ln2 = 2 which implies 2 = e2 which is simply not true.
The derivative of 2^x is 2^x * ln2 so the derivative of 2^cosx * ln2 multiplied by d/dx of cox, which is -sinx so the derivative of the inside function is -sinx * 2^cosx *ln2. As to the final question, using the chain rule, d/dx (2^cosx)^0.5 will equal half of (2^cosx)^-0.5 * -sinx * 2^cosx * ln2
Assuming you mean 'logarithm to the base 'e' ( natural logarithm. On the calculator its symbol is 'ln'. Hence ;ln 2 = 0.69314718....
E is energy and MC2 is matter used to the second power... now you solve it
LN2 i think :D
-2y square exp power -2x-1
900 and divide10 with the power of four solve it = 90
I am not sure what you want to "solve" here. e-x is the same as 1 / ex.
ex = x3 This has two solutions: x = 4.5364... and x = 1.85718... Plot the graph of each and you can see the intersections.