1. ## extraneous roots

8(3)^2x+3 = 24

log3^2x+3 = 3

2x+3 = log3/log3

2x + 3 = 1

x = -1

Is my method correct?

2. 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

$2x = \log_3\frac{21}{8}$

3. 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 ...

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

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

$2x+3 = 1$

$x = -1$

4. 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.

5. you need to work on making the notation of your equations more clear.

parentheses help alot.

6. 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)).

log3^2x+3 = 3
2x+3 = log3/log3

2x + 3 = 1

x = -1

Is my method correct?
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.