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?
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Hello, $\displaystyle \log y=0 \Leftrightarrow y=1$, in any base. so $\displaystyle 6x-3=0$ and x=... ?
x = 1/2 or 0.5 Thanks!
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 $\displaystyle \log_6(x-3) = 0$ , then $\displaystyle x = 4$ if $\displaystyle \log_6{x} - 3 = 0$ , then $\displaystyle x = 6^3$
x-3 should be in the parenthesis, but how do you get 4 then?
do you understand Moo's prior statement? Originally Posted by Moo $\displaystyle \log y=0 \Leftrightarrow y=1$, in any base.
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.
$\displaystyle \log_6(x-3) = 0$ change to an exponential equation ... $\displaystyle 6^0 = x-3$ what is the value of $\displaystyle x$?
ok, so x = 4 Now I have it! Thank you!
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