8(3)^2x+3 = 24
Answer :
log3^2x+3 = 3
2x+3 = log3/log3
2x + 3 = 1
x = -1
Is my method correct?
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
$\displaystyle 2x = \log_3\frac{21}{8}$
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.
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)).
Yes, by taking the logarithm of both sides of $\displaystyle 3^{2x+3}= 3$ you get (2x+3)log 3= log 3 so 2x+ 3= 1.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.