Solve the following expression for x: $log_2(x) = 3$

To solve this expression we need to use the definition of logarithms which states that if: $$log_{a}(b) = c$$ then it is also true that $$a^c = b$$ so using the above definition with a = 2 , b = x and c = 3 we write $$log_{2}(x) = 3$$ then it is also true that $$2^3 = x$$ $$x = 2^3$$ $$x = 8$$

Find the first derivative of the following expression $$f(x) = 2x^{3} $$

so we re write the expression as $$f'(x) = (2x^{3})'$$ the constant goes out of the derivative : $$f'(x) =2 (x^{3})'$$ now using the formula $$(x^{n} )'= n x^{n-1}$$ where n = 3 : $$f'(x) =2(3)(x^{3-1})'$$ This is our answer : $$f'(x) = 6x^{2}$$

Solve the following systems of equations : 3x+5y = 7 3x + 4y = 1

by subtracting the first equation from the second we have : $$3x-3x+5y - 4y = 7 - 1$$ $$y = 6$$ now we substitute the value of y we found in the first equation and we get : $$3x + 5(6) = 7$$ $$3x = 7-30$$ $$3x = -23$$ $$x= -\dfrac{23}{3}$$