How do I simplify the expression into a single log term? log(x^3) + log(1) - 4log(y) + ln(e)
First of all, we know that log(1) = 0, and ln(e) = 1. Also, we will be allowed to move the "4" in "4log(y)" to the exponent of y, as this is one of the logarithmic laws. log(x^3) + 0 - log(y^4) + 1 Now if we want just a single log term, we need to somehow turn this "1" into a log. We can do this by changing it to log10, since this value still equals 1. log(x^3) - log(y^4) + log(10) According to logarithmic laws, when there are addition of logs with the same base, we can multiply their arguments together. When there is subtraction of logs with the same base, we can divide the argument of the value with the negative log. log((x^3 * 10)/(y^4))
Given that tan(𝚹) = 4/3 lies in quadrant I, what would sec^2(𝚹) be?
Since tan(𝚹) = y/x, and sec(𝚹) = hypotenuse/x, we need to determine the hypotenuse value to determine sec(𝚹). We can use the Pythagorean theorem to determine the hypotenuse. I will call the hypotenuse, "h". x^2 + y^2 = h^2 3^2 + 4^2 = h^2 9 + 16 = h^2 25 = h^2 h = +/- 5 Since this is the hypotenuse value, it will always be positive, so we will use +5. sec(𝚹) = h/x sec(𝚹) = 5/3 Since we are looking for sec^2(𝚹), we would square both sides sec^2(𝚹) = 25/9
How do you solve the following Differential Equations? 1) y - 3xy = 25x 2) y = sin(cos(e^(2*x))
1) First, do implicit differentiation, where we are differentiating with respect to x. y' can also represent dy/dx. y' - 3(x*y' + y'*1) = 25 Now expand. y' - 3xy' - 3y = 25 Isolate y' by leaving the terms with y' on the same side in order to factor it out. y' - 3xy' = 25 + 3y y'(1 - 3x) = 25 + 3y y' = (25 + 3y)/(1 - 3x) 2) This involves the chain rule. We take the derivative of the outside, followed by the derivative of the insides. y = cos(tan(e^2*x)) * -sin(e^2*x)) * e^2x * 2 Now I'll simplify it. y = -2cos(tan(e^2*x)) * sin(e^2*x)) * e^2x