Logs

**missyd819** · August 20th 2009, 12:32 PM

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?

**Moo** · August 20th 2009, 12:35 PM

Hello,

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

so $6x-3=0$ and x=... ?

**missyd819** · August 20th 2009, 01:12 PM

x = 1/2 or 0.5

Thanks!

**skeeter** · August 20th 2009, 01:43 PM

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 $x = 6^3$

**missyd819** · August 20th 2009, 01:49 PM

x-3 should be in the parenthesis, but how do you get 4 then?

**skeeter** · August 20th 2009, 01:51 PM

do you understand Moo's prior statement?

Originally Posted by Moo

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

**missyd819** · August 20th 2009, 02:02 PM

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.

**skeeter** · August 20th 2009, 02:11 PM

$\log_6(x-3) = 0$

change to an exponential equation ...

$6^0 = x-3$

what is the value of $x$ ?

**missyd819** · August 20th 2009, 02:19 PM

ok, so x = 4

Now I have it! Thank you!

Thread: Logs