Logarithms

[arli101]

hey, I'm a high school student and I'm currently doing some revision worksheets in an attempt at reviewing logarithms for my finals. However I've been stuck on this particular question for the longest while. I'd appreciate any help that I'm able to get.

here's what I've tried so far:



the question: log (3x-1) + log (3x+1) = log 16

 

[Cubist]

Hi, welcome to the forum and well done for showing your work.



You made a mistake on the 3rd line, hint 3*3 = 9
Also, if log(p) = log(16) then what can you say about the value of p?

 

[JeffM]

Here are two tips that will help you on many such equations.

First tip

[MATH]log_a(b) = log_a(c) \implies b = c.[/MATH]
Notice that this is true no matter what the base a is.

How can you use that in this problem?

[MATH]log(3x - 1) + log(3x + 1) = log(16) \implies\\ log\{(3x - 1)(3x + 1)\} = log(16) \implies \\ \{(3x - 1)(3x + 1)\} = 16 \implies WHAT?[/MATH]Second tip

[MATH]log_a(b) = n \implies a^n = b.[/MATH]
This is particularly helpful when n = 0 because

[MATH]log_a(b) = 0 \implies b = a^0 = 1.[/MATH]
How does that help you in this problem?

[MATH]log(3x - 1) + log(3x + 1) = log(16) \implies\\ log(3x - 1) + log(3x + 1) - log(16) = 0 \implies\\ log \left ( \dfrac{(3x - 1)(3x + 1)}{16} \right ) = 0 \implies\\ \dfrac{(3x - 1)(3x + 1)}{16} = 1 \implies WHAT?[/MATH]

 

[HallsofIvy]

It has already been pointed out that \(\displaystyle log((3x- 1)(3x+ 1))= log(9x^2- 1)= log(16)\).

Now I would NOT write that as \(\displaystyle log(9x^2- 1)- log(16)= log\left(\frac{9x^2- 1}{16}\right)= 0\).

Instead I would just use the fact that the logarithm is "one to one" so that \(\displaystyle 9x^2- 1= 16\).
\(\displaystyle 9x^2= 17\)
\(\displaystyle x^2= \frac{17}{9}\)
\(\displaystyle x= \pm\frac{\sqrt{17}}{3}\).

 

[lookagain]

. . .
\(\displaystyle x= \pm\frac{\sqrt{17}}{3}\).

From the original equation log(3x - 1) + log(3x + 1) = log(16), the correct solution
is \(\displaystyle \ x \ = \ \tfrac{\sqrt{17}}{3}\).

 

[Steven G]

It has already been pointed out that \(\displaystyle log((3x- 1)(3x+ 1))= log(9x^2- 1)= log(16)\).

Now I would NOT write that as \(\displaystyle log(9x^2- 1)- log(16)= log\left(\frac{9x^2- 1}{16}\right)= 0\).

Instead I would just use the fact that the logarithm is "one to one" so that \(\displaystyle 9x^2- 1= 16\).
\(\displaystyle 9x^2= 17\)
\(\displaystyle x^2= \frac{17}{9}\)
\(\displaystyle x= \pm\frac{\sqrt{17}}{3}\).

Dr Halls,
I truly respect you as a mathematician. I just do not know what to say sometimes about your ability to do basic math! Logs do have a restricted domain! I am sure that you know that better than I do.
I am just messing with you. I am always amused at some things that brilliant people say and actually post.
Have a great day!
Steve

 

[HallsofIvy]

I certainly won't say that I know mathematics better than you do, but I do often amaze myself with how careless I can be!

 

[eddy2017]

I am printing the whole thread and stuying it now that i am learning how to do logs.This is amazing. I don't think there is another site like this online. No, none like this!