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?

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.

what exactly is antilog?

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

ANSWER=3

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

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

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

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

let's try that again

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

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

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

To solve the equation Log(X) + Log(X-3) = 1, let's go through the steps again and identify the mistake:

Step 1: Initially, you correctly combined the logarithms using the logarithmic property Log(a) + Log(b) = Log(ab).

So, the equation becomes Log(X(X-3)) = 1.

Step 2: Next, you applied the power rule of logarithms Log(a^b) = b * Log(a) correctly.

Hence, the equation becomes Log(X^2 - 3X) = 1.

Step 3: Instead of subtracting 3X from X^2 directly as you did, we need to combine the two terms under one logarithm by factoring out an X.

So, the equation should be Log(X(X-3)) = 1.

Step 4: You mistakenly simplified Log(X(X-3)) to become Log(-2X). However, X(X-3) ≠ -2X. The equation is still Log(X(X-3)) = 1.

Step 5: To get rid of the logarithm, we can convert the logarithmic equation into an exponential equation.

Rewrite the equation as 10^1 = X(X-3) since 10^1 is equivalent to Log base 10.

Step 6: Simplify this equation to X^2 - 3X = 10.

Now, you can proceed to solve the quadratic equation by subtracting 10 from both sides:

X^2 - 3X - 10 = 0.

Step 7: Factor the quadratic equation:

(X - 5)(X + 2) = 0.

Step 8: Set each factor equal to zero:

X - 5 = 0 or X + 2 = 0.

Solve for X:

X = 5 or X = -2.

After solving the quadratic equation, you find two possible values of X: X = 5 and X = -2.

However, you need to check these solutions in the original logarithmic equation to ensure they are valid.

Plug X = 5 into the equation (Log 5 + Log (5-3) = 1):

Log 5 + Log 2 = 1.

This equation is true, so X = 5 is a valid solution.

However, plug X = -2 into the equation (Log (-2) + Log (-2-3) = 1):

Log(-2) is undefined, so X = -2 is not a valid solution.

Therefore, the only solution to the equation Log(X) + Log(X-3) = 1 is X = 5.