Algebra 2
posted by HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!! .
how do not understand how to do this
Log X + Lob (X3) = 1
I know I do this
Log X(X3)=1
then I do this
Log X^(2)  3X = 1
then I do this
2 Log (X3X) = 1
then
2 Log (2X) = 1
then
(2 Log (2X) = 1)(1/2)
Log (2X) = 1/2
then
(Log (2X) = 1/2)(1/2)
Log X = (1/4)
then
by def
10^ (1/4) = X
I get something under one and calculator tells me I can't do that I think I'm doing something wrong what do I do?

Algebra 2 
bobpursley
Log X + Lob (X3) = 1
I know I do this
Log X(X3)=1
Then take the antilog of each side
x(x3)=0
Now solve it. 
Algebra 2 
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
what exactly is antilog?

Algebra 2 
Reiny
your notation at Log X^(2)  3X = 1 is sloppy
say: Log (X^(2)  3X) = 1
your next line of 2 Log (X3X) = 1
is WRONG
from Log (X^(2)  3X) = 1 by definition of logs
x^2  3x = 10^1
x^2  3x  10 = 0
(x5)(x+2) = 0
x = 5 or x = 2
but in logx, x > 0, so
x = 5
check: if x=5
LS = log5 + log2
= log(5x2)
= log 10
= 1
= RS 
Algebra 2 
Anonymous
ANSWER=3

Algebra 2 
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ok um so
Log (X^(2)  3X) = 1
x^2  3x = 10^1
what allows you to put in the value ten and get ride of the log on the other side? What if the value wasn't one would u just do 10^X?
x^2  3x  10 = 0
(x5)(x+2) = 0
x = 5 or x = 2 
Algebra 2 
Reiny
NO, it is x=5
I verified it
Here is yours:
if x=3
LS= log3 + log1
= log(3x1)
= log 3 which is not equal to the RS of 1 
Algebra 2 
Reiny
LOG_{b<?sub> a = c <> bc = a e.g. log2 8 = 3 <> 23 = 8}

Algebra 2 
Reiny
let's try that again
LOG_{b} a = c <> b^{c} = a
e.g.
log_{2} 8 = 3 <> 2^{3} = 8 
Algebra 2 
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
ok um so
Log (X^(2)  3X) = 1
x^2  3x = 10^1
what allows you to put in the value ten and get ride of the log on the other side? What if the value wasn't one would u just do 10^X?
x^2  3x  10 = 0
(x5)(x+2) = 0
x = 5 or x = 2
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