Question

Solving Exponential and Logarithmic Equations In Exercise, solve for x. In 2x - In(3x - 1) = 0

Answer 1

Answer:

$\frac{-1}{3x^2-x}$

Step-by-step explanation:

1. If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
2. When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)

using these information:

g(x)=ln2x then g'(x)=$\frac{(2x)'}{2x} =\frac{2}{2x}=\frac{1}{x}$

h(x)=In(3x - 1) then h'(x)=$\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}$