# Logarithms

#### Cubist

##### Senior Member
• Subhotosh Khan and JeffM

#### JeffM

##### Elite Member
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]

[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]

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• topsquark

#### HallsofIvy

##### Elite Member
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 $$\displaystyle 9x^2- 1= 16$$.
$$\displaystyle 9x^2= 17$$
$$\displaystyle x^2= \frac{17}{9}$$
$$\displaystyle x= \pm\frac{\sqrt{17}}{3}$$.

• topsquark

#### lookagain

##### Elite Member
. . .
$$\displaystyle x= \pm\frac{\sqrt{17}}{3}$$.

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

#### Jomo

##### Elite Member
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

##### Elite Member
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!

• Otis and Jomo

#### eddy2017

##### Senior Member
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!