Question

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

Answer 1

`(-1)/(3x^2-x)`

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)=
`((2x)')/(2x) =(2)/(2x)=(1)/(x)`

h(x)=In(3x - 1) then h'(x)=
`((3x-1)')/(3x-1) =(3)/(3x-1)</p><p>f'(x)=g'(x) - h'(x) =[tex](1)/(x) - (3)/(3x-1) =(-1)/(3x^2-x)`