8(3)^2x+3 = 24

Answer :

log3^2x+3 = 3

2x+3 = log3/log3

2x + 3 = 1

x = -1

Is my method correct?

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- Jan 2nd 2010, 01:53 PMpurpledinosaurextraneous roots
8(3)^2x+3 = 24

Answer :

log3^2x+3 = 3

2x+3 = log3/log3

2x + 3 = 1

x = -1

Is my method correct? - Jan 2nd 2010, 02:03 PMpickslides
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

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

Taking 3 from both sides gives

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

Now diving both sides by 8 you get

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

Now you need to employ a log

$\displaystyle 2x = \log_3\frac{21}{8}$ - Jan 2nd 2010, 02:05 PMskeeter
- Jan 2nd 2010, 02:09 PMpurpledinosaur
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 PMskeeter
you need to work on making the notation of your equations more clear.

parentheses help alot. - Jan 3rd 2010, 01:11 AMHallsofIvy
Important point: if you do not use LaTex, use

**parentheses**!

"8(3)^2x+ 3" could be interpreted as $\displaystyle 8(3^2)x+ 3$, $\displaystyle 8(3^{2x})+3$ or $\displaystyle 8(3^{2x+3})$.

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

Quote:

Answer :

log3^2x+3 = 3

2x+3 = log3/log3

2x + 3 = 1

x = -1

Is my method correct?

As skeeter pointed out, it is more "fundamental" and easier to argue, from $\displaystyle 3^{2x+3}= 3^1= 3$ that, because exponentials are "one to one" functions ($\displaystyle a^x= a^y$ 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.