Question

In(5x-1)=in(3x+2) extraneous solution

Answer 1

`\ln(5x-1)=\ln(3x+2)`

First note that the left side exists only for
`5x-1>0`, or
`x>\frac15`, and the right side exists only for
`3x+2>0`, or
`x>-\frac23`. Then any solution to this equation is extraneous if we find
`x\le\frac15`.

Write both sides as powers of
`e` to eliminate the logarithms:

`e^(\ln(5x-1))=e^(\ln(3x+2))\implies5x-1=3x+2`

Simplify and solve:

`2x=3\implies x=\frac32`

No extraneous solutions here because the one found is larger than 1/5.