# Solve the logarithm.

• January 22nd 2010, 01:29 PM
bhuang
Solve the logarithm.
$log5(x-1)$+ $log5(x-2)$- $log5(x+6)$=0

the base is 5.
how do you do that on here? i can't seem to understand how to work the math tags button so I can put othedr notations like sigma or the integral sign...

anyways, i got the answer to be x=4.82 or x=-0.83
is that right?
• January 22nd 2010, 02:12 PM
Soroban
Hello, bhuang!

Click on "Quote" to see my code . . .

Quote:

$\log_5(x-1) + \log_5(x-2) - \log_5(x+6) \:=\:0$

i got the answer to be: $x=4.82$ or $x=-0.83$
Is that right? . . . . almost

That negative root is extraneous . . .
Normally, we are not allowed logs of negative quantities.