You are not logged in.
Pages: 1
ahhh
2^2x - 33(2^x) + 32 = 0
5^x^2 (625) = (1/25)^2x
help??:)
or it can go like this:
1) 2²ª 33(2ª) + 32 = 0
²
2) 5ª (625) = ( 1 )²ª
25
The second one is five to the power of "a" squared.....and it equals bracket one over 25 bracket, all to the power of 2a
lol thanks
1) a = 0 is one possibility
2) I don't know what order to do the power in (5^x)^2, or is it 5^(x^2) ?
Last edited by John E. Franklin (2007-09-19 08:44:01)
igloo myrtilles fourmis
Offline
1) a = 0 is one possibility
2) I don't know what order to do the power in (5^x)^2, or is it 5^(x^2) ?
Im not sure either...
I only have written here 5^x^2 .....
(1)
Let x = 2[sup]a[/sup].
Then your equation is x² - 33x + 32 = 0.
(x-32)(x-1) = 0
x= 32 or 1.
Substituting back, 2[sup]a[/sup] is 0 when x=1, and 5 when x=32.
So your two answers are 1 and 5.
Why did the vector cross the road?
It wanted to be normal.
Offline
On the last problem, I tried -2 and it works if
you square the -2 first in the power and then do the 5 up 4.
So (5^(-2^2) )(625) = (1/25)^-4
igloo myrtilles fourmis
Offline
Pages: 1