Question

solve for x, log e (x-a)+logex=1 (e is on the base)

Answer 1

Final Answer :
x :
$x = \frac{a + \sqrt{ {a}^{2} + 4e } }{2}$

Steps :
1)
$\: ln(x - a) + ln(x) = 1 \\ \: \: domain \: = x > a \: \: \: and \: \: x > 0$
$ln(x \times( x - a)) = 1 \\ = > x \times (x - a) = {e}^{1} \\ = > {x}^{2} - ax - e = 0 \\ = > x = \frac{a + \sqrt{ {a}^{2} + 4e } }{2} \: \: \\ or \: x \: = \frac{a - \sqrt{ {a}^{2} + 4e } }{2}$
2) We will reject second part of solution as x > a.
3 ) So, Finally


$x = \frac{a + \sqrt{ {a}^{2} + 4e} }{2}$