# extraneous roots

• Jan 2nd 2010, 01:53 PM
purpledinosaur
extraneous roots
8(3)^2x+3 = 24

log3^2x+3 = 3

2x+3 = log3/log3

2x + 3 = 1

x = -1

Is my method correct?
• Jan 2nd 2010, 02:03 PM
pickslides
Hi purple dinosaur, i'll kick this one off for you.

When solving for x you need to undo some of these operations

You have

$8(3)^{2x}+3 = 24$

Taking 3 from both sides gives

$8(3)^{2x} = 21$
Now diving both sides by 8 you get

$(3)^{2x} = \frac{21}{8}$

Now you need to employ a log

$2x = \log_3\frac{21}{8}$
• Jan 2nd 2010, 02:05 PM
skeeter
Quote:

Originally Posted by purpledinosaur
8(3)^2x+3 = 24

log3^2x+3 = 3

2x+3 = log3/log3

2x + 3 = 1

x = -1

Is my method correct?

it works.

would be easier to do this ...

$8 \cdot 3^{2x+3} = 24$

$3^{2x+3} = 3$

$2x+3 = 1$

$x = -1$
• Jan 2nd 2010, 02:09 PM
purpledinosaur
my apologies to you, i think u started answering my question when i posted the original question. I stared it for about 20 mins, and didn't understand it but when i posted the question i sort of had an epiphany, and i edited my post. Thank you for clearing it for me. And for the explanation, much appreciated. (Hi)
• Jan 2nd 2010, 02:21 PM
skeeter
you need to work on making the notation of your equations more clear.

parentheses help alot.
• Jan 3rd 2010, 01:11 AM
HallsofIvy
Quote:

Originally Posted by purpledinosaur
8(3)^2x+3 = 24

Important point: if you do not use LaTex, use parentheses!
"8(3)^2x+ 3" could be interpreted as $8(3^2)x+ 3$, $8(3^{2x})+3$ or $8(3^{2x+3})$.

From what you do below, you appear to mean 8(3^(2x+3)).

Quote:

Yes, by taking the logarithm of both sides of $3^{2x+3}= 3$ you get (2x+3)log 3= log 3 so 2x+ 3= 1.
As skeeter pointed out, it is more "fundamental" and easier to argue, from $3^{2x+3}= 3^1= 3$ that, because exponentials are "one to one" functions ( $a^x= a^y$ if and only if x= y) that we must have 2x+3= 1. From that, 2x= 1- 3= -2 and so x= -2/2= -1.