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 ^ (1-x)

- 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)(y-3) = 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 ^ (1-x)

log(3/5)^x = log7 ^ (1-x)

xlog(3/5) = (1-x)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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