Posted by mia on Monday, December 1, 2008 at 8:27pm.
how do i solve these?
9^(x)=1/3
2^(2x)+ 2^(x) 12 = 0
(3/5)^x = 7 ^ (1x)

math precalculus  Reiny, Monday, December 1, 2008 at 8:36pm
9^(x)=1/3
(3^2)^(x) = 3^1
3^(2x) = 3^1
so 2x = 1
x = 1/2
2^(2x)+ 2^(x) 12 = 0
(2^x)^2 + 2^x  12 = 0
let y = 2^x
y^2 + y  12 = 0
(y+4)(y3) = 0
y = 4 or y = 3
2^x = 4, no solution , or
2^x = 3  x = log3/log2

math precalculus  Reiny, Monday, December 1, 2008 at 8:40pm
last one:
take logs of both sides
(3/5)^x = 7 ^ (1x)
log(3/5)^x = log7 ^ (1x)
xlog(3/5) = (1x)Log 7
xlog .6 = log 7  xlog 7
xlog .6 + xlog 7 = log 7
x(log .6 + log 7) = log 7
x = log 7/(log.6 + log7)

math precalculus  Angie, Tuesday, December 2, 2008 at 12:32am
1/6

rwil ohsuycip  rwil ohsuycip, Saturday, January 17, 2009 at 4:43am
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