If 2^4x = 16, then x = ?
2^(4x) can be rewritten as (2^4)^x which simplifies to 16^x. Taking log_16 of both sides of the equation 16^x = 16 gives us that x = 1, which we can also just find out by plugging in numbers. Ans: x = 1
What is the limit as x approaches infinity of (4x^2-2x)/(1-3x^2)?
The limit can be found using the L'Hospitals Rule which states that if a limit is in the form 0/0 or infinity/infinity then its numerator and denominator can be replaced with the derivatives of each. Taking the derivative once on both gives us lim x->infinity of (8x - 2)/(-6x) but taking it once more since it is still in that form gives the answer: 8/-6 or simplified --> -4/3
What is a feature in the C++ programming language that is not available in the C programming language?
The ability to create classes and the STL library.