here's what I've tried so far:

the question:

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

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- 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]

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

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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

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