Posted by **HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!** on Wednesday, February 25, 2009 at 8:08pm.

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 -
**bobpursley**, Wednesday, February 25, 2009 at 8:11pm
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 -
**HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!**, Wednesday, February 25, 2009 at 8:12pm
what exactly is antilog?

- Algebra 2 -
**Reiny**, Wednesday, February 25, 2009 at 8:16pm
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 -
**Anonymous**, Wednesday, February 25, 2009 at 8:28pm
ANSWER=3

- Algebra 2 -
**HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!**, Wednesday, February 25, 2009 at 8:31pm
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 -
**Reiny**, Wednesday, February 25, 2009 at 8:33pm
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**, Wednesday, February 25, 2009 at 8:38pm
LOG_{b<?sub> a = c <----> bc = a
e.g.
log2 8 = 3 <---> 23 = 8
}

- Algebra 2 -
**Reiny**, Wednesday, February 25, 2009 at 8:39pm
let's try that again

LOG_{b} a = c <----> b^{c} = a

e.g.

log_{2} 8 = 3 <---> 2^{3} = 8

- Algebra 2 -
**HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!**, Wednesday, February 25, 2009 at 8:41pm
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

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