# Algebra 2

posted by on .

how do not understand how to do this

Log X + Lob (X-3) = 1

I know I do this

Log X(X-3)=1

then I do this

Log X^(2) - 3X = 1

then I do this

2 Log (X-3X) = 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 - ,

Log X + Lob (X-3) = 1

I know I do this

Log X(X-3)=1

Then take the antilog of each side
x(x-3)=0
Now solve it.

• Algebra 2 - ,

what exactly is antilog?

• Algebra 2 - ,

your notation at Log X^(2) - 3X = 1 is sloppy

say: Log (X^(2) - 3X) = 1

your next line of 2 Log (X-3X) = 1
is WRONG

from Log (X^(2) - 3X) = 1 by definition of logs
x^2 - 3x = 10^1
x^2 - 3x - 10 = 0
(x-5)(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 - ,

• Algebra 2 - ,

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
(x-5)(x+2) = 0
x = 5 or x = -2

• Algebra 2 - ,

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 - ,

LOGb<?sub> a = c <----> bc = a

e.g.
log2 8 = 3 <---> 23 = 8

• Algebra 2 - ,

let's try that again

LOGb a = c <----> bc = a

e.g.
log2 8 = 3 <---> 23 = 8

• Algebra 2 - ,

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
(x-5)(x+2) = 0
x = 5 or x = -2