# Logs

• Aug 20th 2009, 12:32 PM
missyd819
Logs
If I have Log 6 x-3 = 0 (the 6 should be a small 6 attached to the log) It is not working for some reason!

How can I solve for x?
• Aug 20th 2009, 12:35 PM
Moo
Hello,

$\log y=0 \Leftrightarrow y=1$, in any base.

so $6x-3=0$ and x=... ?
• Aug 20th 2009, 01:12 PM
missyd819
x = 1/2 or 0.5

Thanks!
• Aug 20th 2009, 01:43 PM
skeeter
Quote:

Originally Posted by missyd819
If I have Log 6 x-3 = 0 (the 6 should be a small 6 attached to the log) It is not working for some reason!

How can I solve for x?

your notation needs some work ...

if $\log_6(x-3) = 0$ , then $x = 4$

if $\log_6{x} - 3 = 0$ , then $x = 6^3$
• Aug 20th 2009, 01:49 PM
missyd819
x-3 should be in the parenthesis, but how do you get 4 then?
• Aug 20th 2009, 01:51 PM
skeeter
do you understand Moo's prior statement?

Quote:

Originally Posted by Moo

$\log y=0 \Leftrightarrow y=1$, in any base.

• Aug 20th 2009, 02:02 PM
missyd819
No because she had 6x - 3 = 0, so I though x = 1/2. I have no idea how to even start to get close to 4.
• Aug 20th 2009, 02:11 PM
skeeter
$\log_6(x-3) = 0$

change to an exponential equation ...

$6^0 = x-3$

what is the value of $x$?
• Aug 20th 2009, 02:19 PM
missyd819
ok, so x = 4

Now I have it! Thank you!