how do you do this problem?
4^x+2 - 4^x = 15
its got me stumped. are logarithms supposed to be involved?
2007-09-17 21:41:08 · 7 answers · asked by nasgmx,zncv xc 2 in Science & Mathematics ➔ Mathematics
opps here let me make the problem a little clearer:
4^(x+2) - 4^x = 15
2007-09-17 21:51:05 · update #1
4^(x+2)-4^x=15
4^x(4^2-1)=15
4^x*15=15
4^x=1=4^0
x=0 ans
2007-09-17 22:01:41 · answer #1 · answered by alpha 7 · 0⤊ 0⤋
Its quite tricky to think of log to be ur process on the problem. however its not....
4^(x+2) - 4^x = 15
4^x * 4^2 - 4^x = 15........Multiplying expression w/ the same base, add the exp.
16*4^x - 4^x = 15
4^x(16 - 1) = 15..............common monomial factor is 4^x
15*4^x = 15
(15*4^x)/15 = 15/15
4^x = 1
4^x = 4^0........................ any number with exponent 0 is 1
x = 0
Check:
4^(x+2) - 4^x = 15
x = 0
4^(0+2) - 4^0 = 15 ?
4^2 - 4^0 = 15 ?
16 - 1 = 15 ?
15 = 15 check....
2007-09-17 23:21:54 · answer #2 · answered by Jhun2x 1 · 0⤊ 1⤋
Taking logs directly is a little tricky because you will not be able to separte the left hand side easily.
Try multiplying through by 4^-x so you get down to one x term only.
This will give
4^2-1=15*4^-x
15=15*4^-x
1=4^-x
therefore x=0
As a check, 4^2-4^0=15.
2007-09-17 21:57:36 · answer #3 · answered by Random Maths Guy 2 · 0⤊ 0⤋
Factor out 4^x to get
4^x * (4²-1)=15
4^x * 15 = 15
4^x = 1 and x=0
Doug
2007-09-17 22:04:26 · answer #4 · answered by doug_donaghue 7 · 0⤊ 0⤋
we can notice immediately that 4^(x+2)=(4^x)*(4^2)=16(4^x)
then 4^x+2 - 4^x=(4^x)*(16-1)-15
4^x=1, x=0
2007-09-17 22:12:37 · answer #5 · answered by marcus101 2 · 0⤊ 0⤋
( 4^x )( 4 ² ) - 4^x = 15
( 4^x ) ( 16 - 1 ) = 15
( 4^x ) (15) = 15
4^x = 1
x = 0
2007-09-20 07:22:39 · answer #6 · answered by Como 7 · 1⤊ 0⤋
4^(x + 2) - 4^x = 15
4^x (4^2 - 1) = 15
take 4^x coomon...
4^x (16 - 1) = 15
4^x (15) = 15
15 goes down...
4^x = 15/15
4^x = 1
this is the answer...
2007-09-17 21:55:47 · answer #7 · answered by bali... 2 · 0⤊ 1⤋