# Math Help - Equations involving variables and Logarithms

1. ## 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.

2. ## Re: Equations involving variables and Logarithms

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

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

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

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

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

3. ## Re: Equations involving variables and Logarithms

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

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

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

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

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

4. ## Re: Equations involving variables and Logarithms

You have ended up with $\displaystyle \log{\left[\frac{3(x - 1)}{2x}\right]} = 0$. How do you undo a logarithm?

5. ## Re: Equations involving variables and Logarithms

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 $\log(1)=~?$

log (1) = 0

7. ## 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.

8. ## Re: Equations involving variables and Logarithms

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

Can you go from here?

9. ## Re: Equations involving variables and Logarithms

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

Can you go from here?
no i can not.

10. ## Re: Equations involving variables and Logarithms

Originally Posted by Prove It
In that case, simplify the RHS of your original equation to get $\displaystyle \log{(2x)} = \log{[3(x - 1)]}$.
Can you go from here?
Originally Posted by bkbowser
no i can not.
Do you understand that if $\log(X)=\log(Y)\text{ then }X=Y~?$

11. ## Re: Equations involving variables and Logarithms

Originally Posted by Plato
Do you understand that if $\log(X)=\log(Y)\text{ then }X=Y~?$
apparently not.

3x-3 = 2x
3x-2x = 3
x = 3

12. ## Re: Equations involving variables and Logarithms

Originally Posted by bkbowser
apparently not.
3x-3 = 2x
3x-2x = 3
x = 3
Now you have done it correctly. Good!