# Equations involving variables and Logarithms

• Jul 5th 2011, 10:57 AM
bkbowser
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.
• Jul 5th 2011, 11:08 AM
e^(i*pi)
Re: Equations involving variables and Logarithms
You can use the laws of logarithms to simplify your equations into one containing a single log

$\displaystyle \log_{b}(a) + \log_b(c) = \log_b(ac)$

$\displaystyle \log_{b}(a) - \log_b(c) = \log_b \left(\dfrac{a}{c}\right)$

$\displaystyle \log_b(a) = c \Leftrightarrow a = b^c$

If I take your first equation we can use the first law to say that $\displaystyle \log(2x) = \log(3) + \log(x-1) \rightarrow \log \left(\dfrac{3(x-1)}{2x}\right) = 0$.
• Jul 5th 2011, 11:19 AM
bkbowser
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

$\displaystyle \log_{b}(a) + \log_b(c) = \log_b(ac)$

$\displaystyle \log_{b}(a) - \log_b(c) = \log_b \left(\dfrac{a}{c}\right)$

$\displaystyle \log_b(a) = c \Leftrightarrow a = b^c$

If I take your first equation we can use the first law to say that $\displaystyle \log(2x) = \log(3) + \log(x-1) \rightarrow \log \left(\dfrac{3(x-1)}{2x}\right) = 0$.

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).
• Jul 5th 2011, 11:27 AM
Prove It
Re: Equations involving variables and Logarithms
You have ended up with $\displaystyle \displaystyle \log{\left[\frac{3(x - 1)}{2x}\right]} = 0$. How do you undo a logarithm?
• Jul 5th 2011, 11:28 AM
Plato
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 $\displaystyle \log(1)=~?$
• Jul 5th 2011, 11:34 AM
bkbowser
Re: Equations involving variables and Logarithms
log (1) = 0
• Jul 5th 2011, 11:38 AM
bkbowser
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.
• Jul 5th 2011, 11:47 AM
Prove It
Re: Equations involving variables and Logarithms
In that case, simplify the RHS of your original equation to get $\displaystyle \displaystyle \log{(2x)} = \log{[3(x - 1)]}$.

Can you go from here?
• Jul 5th 2011, 12:18 PM
bkbowser
Re: Equations involving variables and Logarithms
Quote:

Originally Posted by Prove It
In that case, simplify the RHS of your original equation to get $\displaystyle \displaystyle \log{(2x)} = \log{[3(x - 1)]}$.

Can you go from here?

no i can not.
• Jul 5th 2011, 12:32 PM
Plato
Re: Equations involving variables and Logarithms
Quote:

Originally Posted by Prove It
In that case, simplify the RHS of your original equation to get $\displaystyle \displaystyle \log{(2x)} = \log{[3(x - 1)]}$.
Can you go from here?

Quote:

Originally Posted by bkbowser
no i can not.

Do you understand that if $\displaystyle \log(X)=\log(Y)\text{ then }X=Y~?$
• Jul 5th 2011, 12:38 PM
bkbowser
Re: Equations involving variables and Logarithms
Quote:

Originally Posted by Plato
Do you understand that if $\displaystyle \log(X)=\log(Y)\text{ then }X=Y~?$

apparently not.

3x-3 = 2x
3x-2x = 3
x = 3
• Jul 5th 2011, 12:40 PM
Plato
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!