27^(2x) = 9^(x-3)
solve for x
thnx a lot for ur help!
2006-08-26 12:32:31 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I'm not sure what the up arrow is in your equation. Could you let us know for sure? I will assume you mean multipication here.
27 (times) (2x) = 9 (times)(x-3)
54x = 9x - 27
Subtracting 9x from both sides, 45x = -27
dividing both sides by 45, x = -27/45 = -3/5 = -0.6
If up arrow is an exponent, then Flit (the previous answer) is correct.
2006-08-26 12:41:37 · answer #1 · answered by F. Frederick Skitty 7 · 0⤊ 1⤋
If that is the only equation you have then you will have to substitute every single number for x and see which ones will give you a balanced equation. There may be a quick method of doing it, but either I missed that lesson or have already forgotten it. Usually, this type of question has 2 or more equations which you can manipulate to give you x on 1 side and a value on the other. If not, try (27^(2x))/(0^(x-3))=1 and see if that gives a valid answer.
Or I can stop answering math questions. Shees am I thick. Just call me Forgetful Jones. Flit deserves best but do you understand what happened so you can do the next one that is similar?
2006-08-26 19:44:23 · answer #2 · answered by St N 7 · 0⤊ 1⤋
get both sides to be the same bases.
27^(2x) = 9^(x-3)
3^ 3= 27
3^ 2 = 9
therefore, switch both sides to base 3.
3^ (2x*3) = 3^ (2(x-3))
simplify the exponent
3^(6x) = 3^(2x-6)
Since the bases are the same, the statement will read this equals this. So, the exponents must be equal, right?
6x = 2x-6
simplify
4x = -6
x = -3/2
check it
27^ (-3/2 * 2) = 9^ ((-3/2)-3)
27^(-3) = 9^(-9/2)
plug it into the calculator!
It should be 5.0805...* 10 ^ -5 = 5.0805...* 10 ^ -5
...and so it's correct.
And you are very welcome.
2006-08-26 19:40:19 · answer #3 · answered by flit 4 · 3⤊ 0⤋
Okay...
27 = 3^3
9 = 3^2
3^3(2x) = 3^2(x - 3)
3^6x = 3^(2x - 6)
That means exponents are equal...
6x = 2x -6
4x = -6
x = -6/4
x = -3/2
There you go. Hope you figured out where you messed up. :-)
2006-08-26 22:09:57 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Start by turning both sides of the equation into indices - powers of 3.
27 = 3^3
9 = 3^2
So 3^(6x) = 3^(2(x-3))
Then you can say 6x = 2x - 6
and go from here
2006-08-26 19:43:03 · answer #5 · answered by Orinoco 7 · 1⤊ 0⤋
I'm a little rusty at laws of logs, but here goes.
Take the natural log of both sides:
ln27^2x=ln9^(x-3)
2x ln27 = (x-3) ln9
2x/(x-3)=ln9/ln27 = 2/3
2x=2/3(x-3)
6x=2x-6
4x=-6
x=-6/4
x=-3/2
2006-08-26 19:50:57 · answer #6 · answered by Wicked Mickey 4 · 0⤊ 0⤋
2x log(27)=(x-3) log 9
2x/(x-3)=log(9)/log(27)=2/3
2x/(x-3)=2/3 [multiply by (x-3)]
2x=2/3 x -2
4/3x=-2
x=-6/4=-3/2
no time to check this, and I went real fast, maybe its right and maybe it will give you some ideas even if its wrong
2006-08-26 19:44:27 · answer #7 · answered by enginerd 6 · 1⤊ 0⤋
Excellent answer and explanation, Flit!
2006-08-26 19:50:48 · answer #8 · answered by mozart 3 · 0⤊ 0⤋
I could help you if I knew what that ^ stands for.
They didn't have that when I took Algebra 40 years ago.
2006-08-26 19:40:03 · answer #9 · answered by mia2kl2002 7 · 0⤊ 1⤋
sorry--I'd like to help but I'm an English major..
2006-08-26 19:38:50 · answer #10 · answered by Lisa J 3 · 0⤊ 1⤋