See the equation.
\[3^{x}(3^{x+2})\]
=9
Are you solving for X?
Yes :)
Let's see, give me a sec. I think there will be some logarithms in here
Alright, thanks!
Ok. so distributing you can see we would get \[3^{2x+2}=9\] We can rewrite 9 as \[3^{2}\]
So we would get \[3^{2x+2}=3^{2}\] then take the log base 3 of both sides and then \[2x+2=2\]
Then you can solve for x
But don't you have to multiply the 3's?
Oh wait. You multiply its squared to the x?
You would take the logbase 3 of both sides \[\log_{3} ( 3^{2x+2}) = \log_{3} (3^{2})\]
leaving you with only exponents You could also maybe see it straight up as well. \[3^{2x+2} = 9 \]
what is 9 it is 3 squared, so you would know x must be 0
I don't see how you get rid of the other three without having to make it into a \[3 ^{x^{2}}\]
Have you ever worked with logarithms? Maybe I am taking the wrong approach ^_^;
I guess you know this much we can write your equation like \[3^{2x+2}= 3^{2}\]
No, I'm in geometry hehe. My teacher didn't even tell us that this topic was logarithms!
Haha. Well in this problem we don't necessarily need them to solve. We see we can write the expression as \[3^{2x+2}= 3^{2}\] each side with the same base so It is clear the exponents on the left must add up to what we have on the right, in our case 2, so x must be 0!
\[3^{x}(3^{x+2)}=9\]
Where do you get the 2x?
Rules of exponents : if we are multiplying like terms we add exponents \[X \times X = X^{2} or X^{1+1}= X^{2}\]
Okay
so what I did in your example is \[3^{x+2+x}\]
combine like terms to get \[3^{2x+2}\]
I actually meant one, and put two by accident! sorry >_< So it's \[3^{2x+1}=9\]
so x=4?
X = 1/2 in this one
Look at it like this \[3^(2x+1) = 3^{2}\]
Whoops, should be: \[3^{2x+1} = 3^{2}\]
We know \[3^{2} = 9\]
Oh, I see my mistake! I forgot to make the 9 into 3^2
Thank you ^_^!
Yeah it is a bit more clear that way :D No problem anytime
I have more questions about this, and I was hoping if you could answer some of them? I have 9 I don't know out of 30
Ok sure!
Thank you so much!
Not a problem. Hopefully I can be of some help.
My first fan! :D lol
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