Question

Multiple choice - what is(are) the solution(s) of log 5x+log(x-1)=2? a.-4,5 b.4 c.5 d.4,5

Answer 1

\[\log5x(x-1)=2\]
\[10^2=5x^2-5x\]
\[x^2-x-20=0\]
\[(x-5)(x+4)=0\]
\[x=5 or x-4\]

Answer 2

c.5

Answer 3

Discard -4 because a negative argument is illegal. So x = 5

Answer 4

log (5x) + log (x-1) =2
log[(5x)(x-1)] = 2
Assuming this is a base 10 log

10^2 = [(5x)(x-1)]
100 = 5x^2 - 5x
5x^2 - 5x - 100 = 0
x^2 - x - 20 = 0
(x - 5) (x + 4) = 0
x = 5 or x =- 4
x = -4 is an extraneious root because it produces the log of a negative number

x = 5 --> Answer