Equations involving variables and Logarithms

I have;

log 2x = log 3+ log (x-1)

and

ln x + ln (x-2) - ln (x+4) = 0

I can not figure out how to separate the variable from the constants.

a\ log 2x = log 3+ log (x-1)

b\ log 3+ log (x-1) - log 2x

c\ (3x-3)/2x

d\ log 3/2 (1-(1/x))

c\ log 3/2 + log (1-(1/x))

this approach doesn't seem to be going any place.

Re: Equations involving variables and Logarithms

You can use the laws of logarithms to simplify your equations into one containing a single log

If I take your first equation we can use the first law to say that .

Re: Equations involving variables and Logarithms

Quote:

Originally Posted by

**e^(i*pi)** You can use the laws of logarithms to simplify your equations into one containing a single log

If I take your first equation we can use the first law to say that

.

I do not see how this helps me at all. step c\ in my post is exactly what you conclude with. which is not the answer (x=3).

Re: Equations involving variables and Logarithms

You have ended up with . How do you undo a logarithm?

Re: Equations involving variables and Logarithms

Quote:

Originally Posted by

**bkbowser** I do not see how this helps me at all. step c\ in my post is exactly what you conclude with. which is not the answer (x=3).

What does

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

i think i'm trying to get a sort of zero (the input needs to be 1 not zero) of the rational function inside of the logarithm. clearly;

(3x-3)/2x=1 when x = 3.

but i can't come up with a generic way to evaluate for x with out just randomly inserting numbers or using a graphing program.

i'm sorry if that wasn't as obvious as i thought it was in the first place or if i am incorrect and am doing the whole thing wrong.

Re: Equations involving variables and Logarithms

In that case, simplify the RHS of your original equation to get .

Can you go from here?

Re: Equations involving variables and Logarithms

Quote:

Originally Posted by

**Prove It** In that case, simplify the RHS of your original equation to get

.

Can you go from here?

no i can not.

Re: Equations involving variables and Logarithms

Quote:

Originally Posted by

**Prove It** In that case, simplify the RHS of your original equation to get

.

Can you go from here?

Quote:

Originally Posted by

**bkbowser** no i can not.

Do you understand that if

Re: Equations involving variables and Logarithms

Quote:

Originally Posted by

**Plato** Do you understand that if

apparently not.

3x-3 = 2x

3x-2x = 3

x = 3

Re: Equations involving variables and Logarithms

Quote:

Originally Posted by

**bkbowser** apparently not.

3x-3 = 2x

3x-2x = 3

x = 3

Now you have done it correctly. Good!