here's what I've tried so far:

the question:

**log (3x-1) + log (3x+1) = log 16**

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter arli101
- Start date

here's what I've tried so far:

the question:

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]

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]

Last edited by a moderator:

- Joined
- Jan 27, 2012

- Messages
- 7,602

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 \(\displaystyle 9x^2- 1= 16\).

\(\displaystyle 9x^2= 17\)

\(\displaystyle x^2= \frac{17}{9}\)

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

. . .

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

From the original equation log(3x - 1) + log(3x + 1) = log(16), the correct solution

is \(\displaystyle \ x \ = \ \tfrac{\sqrt{17}}{3}\).

- Joined
- Dec 30, 2014

- Messages
- 10,718

Dr Halls,

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 \(\displaystyle 9x^2- 1= 16\).

\(\displaystyle 9x^2= 17\)

\(\displaystyle x^2= \frac{17}{9}\)

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

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

- Joined
- Jan 27, 2012

- Messages
- 7,602