Question

Solve loge (x-a) + loge x = 1.​

Answer 1

Answer:

So, solution is,

x = (a/2)+(1/2)(√a^2+4e) and (a/2)-(1/2)(√a^2+4e)

Step-by-step explanation:

Log(x-a) +logx = 1

So, log[(x(x-a)] = 1

So, x(x-a) =e

So, now equation is, x^2 -ax -e =0

So, solution using formula is,

X= (a+-√(a^2+4e))/2

So, solution is,

x = (a/2)+(1/2)(√a^2+4e) and (a/2)-(1/2)(√a^2+4e)