Equations involving variables and Logarithms

**bkbowser** · July 5th 2011, 10:57 AM

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.

**e^(i*pi)** · July 5th 2011, 11:08 AM

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 $\log(2x) = \log(3) + \log(x-1) \rightarrow \log \left(\dfrac{3(x-1)}{2x}\right) = 0$ .

**bkbowser** · July 5th 2011, 11:19 AM

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).

**Prove It** · July 5th 2011, 11:27 AM

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

**Plato** · July 5th 2011, 11:28 AM

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 $\log(1)=~?$

**bkbowser** · July 5th 2011, 11:34 AM

log (1) = 0

**bkbowser** · July 5th 2011, 11:38 AM

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.

**Prove It** · July 5th 2011, 11:47 AM

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

Can you go from here?

**bkbowser** · July 5th 2011, 12:18 PM

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.

**Plato** · July 5th 2011, 12:32 PM

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 $\log(X)=\log(Y)\text{ then }X=Y~?$

**bkbowser** · July 5th 2011, 12:38 PM

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

**Plato** · July 5th 2011, 12:40 PM

Originally Posted by bkbowser

apparently not.
3x-3 = 2x
3x-2x = 3
x = 3

Now you have done it correctly. Good!

Thread: Equations involving variables and Logarithms

LinkBack

Thread Tools

Display

Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Re: Equations involving variables and Logarithms

Similar Math Help Forum Discussions

Solve equation involving logarithms

Derivatives involving logarithms

Functions involving powers of logarithms

Couple questions involving continuity and logarithms....

Inverse Function Problem (involving logarithms)