# Evaluating Logarithmics

• Dec 9th 2007, 04:37 PM
Macleef
Evaluating Logarithmics
Solve

$\displaystyle 27^{x} (9^{2x - 1}) = 3^{x + 4}$

My work:

$\displaystyle 3^{3x} (3^{4x-2}) = 3^{x+4}$

$\displaystyle (3x)(4x-2) = x+4$

$\displaystyle 12x^2 - 6x - x - 4 = 0$

$\displaystyle 12x^2 - 7x - 4 = 0$

Now I'm stuck, I can't find any factors that has the product of 48 and the sum of -7.

• Dec 9th 2007, 04:41 PM
colby2152
Quote:

Originally Posted by Macleef
Solve

$\displaystyle 27^{x} (9^{2x - 1}) = 3^{x + 4}$

Divide by $\displaystyle 3^x$
27/3 = 9, $\displaystyle 3^{x + 4} = 3^x 3^4$

$\displaystyle 9^x(9^{2x - 1})=3^4$

Combine like bases of nine...

$\displaystyle 9^{3x-1}=3^4$

Nine equals three squared...

$\displaystyle 3^{6x-2}=3^4$

Now, take log based three of both sides...

$\displaystyle 6x-2=4$
$\displaystyle x=1$
• Dec 9th 2007, 04:42 PM
Jhevon
Quote:

Originally Posted by Macleef
Solve

$\displaystyle 27^{x} (9^{2x - 1}) = 3^{x + 4}$

My work:

$\displaystyle 3^{3x} (3^{4x-2}) = 3^{x+4}$

$\displaystyle \color{red}(3x)(4x-2) = x+4$

...

that is wrong. $\displaystyle x^a \cdot x^b = x^{a {\color {red} +} b}$ NOT $\displaystyle x^{ab}$
• Dec 9th 2007, 04:48 PM
Macleef
Quote:

Originally Posted by Jhevon
that is wrong. $\displaystyle x^a \cdot x^b = x^{a {\color {red} +} b}$ NOT $\displaystyle x^{ab}$

I don't understand what you wrote...
• Dec 9th 2007, 04:49 PM
Macleef
Quote:

Originally Posted by colby2152
Divide by $\displaystyle 3^x$
27/3 = 9, $\displaystyle 3^{x + 4} = 3^x 3^4$

$\displaystyle 9^x(9^{2x - 1})=3^4$

Combine like bases of nine...

$\displaystyle 9^{3x-1}=3^4$

Nine equals three squared...

$\displaystyle 3^{6x-2}=3^4$

Now, take log based three of both sides...

$\displaystyle 6x-2=4$
$\displaystyle x=1$

Why is it common base of 9? I don't think you can get 27 from any exponents of 9?
• Dec 9th 2007, 04:52 PM
colby2152
Quote:

Originally Posted by Macleef
Why is it common base of 9? I don't think you can get 27 from any exponents of 9?

$\displaystyle 9^x9^{(2x-1)} = 9^{(3x-1)}$

That is what I was referring to. As far as 27, that is equal to 3^3
• Dec 9th 2007, 04:55 PM
Macleef
Quote:

Originally Posted by colby2152
$\displaystyle 9^x9^{(2x-1)} = 9^{(3x-1)}$

That is what I was referring to. As far as 27, that is equal to 3^3

What happens to the rest of 27? I still don't know how you get a common base of 9... (I know $\displaystyle 3^{3}$ ) then wouldn't it be $\displaystyle (3)9^{x}$ instead of just $\displaystyle 9^{x}$?
• Dec 9th 2007, 05:26 PM
colby2152
Quote:

Originally Posted by Macleef
What happens to the rest of 27? I still don't know how you get a common base of 9... (I know $\displaystyle 3^{3}$ ) then wouldn't it be $\displaystyle (3)9^{x}$ instead of just $\displaystyle 9^{x}$?

It starts out with 27^x. 27=3^3, so 27^x = 3^3x
• Dec 9th 2007, 05:34 PM
Macleef
Quote:

Originally Posted by colby2152
Divide by $\displaystyle 3^x$
27/3 = 9, $\displaystyle 3^{x + 4} = 3^x 3^4$

$\displaystyle 9^x(9^{2x - 1})=3^4$

Combine like bases of nine...

$\displaystyle 9^{3x-1}=3^4$

Nine equals three squared...

$\displaystyle 3^{6x-2}=3^4$

Now, take log based three of both sides...

$\displaystyle 6x-2=4$
$\displaystyle x=1$

Okay, I finally get why it's $\displaystyle 9^{x}$ ...

Now, one more question, why is it just $\displaystyle 3^{4}$? What happened to to the "x" from $\displaystyle 3^{x + 4}$?
• Dec 9th 2007, 05:35 PM
colby2152
Quote:

Originally Posted by Macleef
Okay, I finally get why it's $\displaystyle 9^{x}$ ...

Now, one more question, why is it just $\displaystyle 3^{4}$? What happened to to the "x" from $\displaystyle 3^{x + 4}$?

$\displaystyle 3^{x+4}=3^x3^4$
• Dec 9th 2007, 05:38 PM
Macleef
Quote:

Originally Posted by colby2152
$\displaystyle 3^{x+4}=3^x3^4$

what happened to $\displaystyle 3^{x}$?? I don't see it in your work...?? I don't see any common factoring if that's what you did...
• Dec 9th 2007, 05:39 PM
topsquark
Quote:

Originally Posted by Macleef
what happened to $\displaystyle 3^{x}$?? I don't see it in your work...?? I don't see any common factoring if that's what you did...

For any nonzero a:
$\displaystyle a^ma^n = a^{m + n}$

-Dan
• Dec 9th 2007, 05:45 PM
Macleef
I still don't get it...but thanks for helping me anyways...
• Dec 9th 2007, 06:12 PM
topsquark
Quote:

Originally Posted by Macleef
I still don't get it...but thanks for helping me anyways...

Let's try an example:

Let a = 3, m = 4, and n = 2. I'm going to calculate $\displaystyle 3^4 \cdot 3^2$.

$\displaystyle 3^4 \cdot 3^2 = (3 \cdot 3 \cdot 3 \cdot 3) \cdot (3 \cdot 3) = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^6$

Our rule states that $\displaystyle a^ma^n = a^{m + n}$ so
$\displaystyle 3^4 \cdot 3^2 = 3^{4 + 2} = 3^6$ (check!)

Look at both ways of doing the problem and see if it helps you understand where this rule comes from. It's all about counting out the number of factors of "a" that you have.

-Dan