x +2 is all sqrd.
a. x> -10
b. x<-10
c. x>10
d. x<10
2007-04-23 09:41:18 · 5 answers · asked by elona 1 in Science & Mathematics ➔ Mathematics
Change the equation.
2^(6*x) < 2^5(x+2)
then solve 6x < 5x + 10
x < 10, answer D
2007-04-23 09:48:11 · answer #1 · answered by delilah 2 · 0⤊ 0⤋
If 64^x < 32^(x+2)
then (2^6)^x < (2^5)^(x + 2)
2^6x < 2^(5x + 10)
6x < 5x + 10
x < 10
2007-04-23 16:51:58 · answer #2 · answered by Helmut 7 · 2⤊ 0⤋
2^6x<2^5(x+2) so;
6x<5x+10
x<10 the ans is d
2007-04-23 16:47:05 · answer #3 · answered by ... 2 · 0⤊ 0⤋
Since log is a monotonic function, you can take logs of both sides and still maintain the equality sign.
x*log(64) < (x+2)*log(32)
x*(log64 - log32) < 2*log32
x < 2*log32 / log2
x < 10
Note: It does not matter which log you use as long as you're consistent
2007-04-23 16:47:06 · answer #4 · answered by Dr D 7 · 1⤊ 1⤋
So this is (32)^(x+2)^2...I doubt it..try rewriting correctly.
2007-04-23 16:44:04 · answer #5 · answered by victoria 5 · 0⤊ 0⤋