# math- precalculus

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how do i solve these?

9^(-x)=1/3

2^(2x)+ 2^(x) -12 = 0

(3/5)^x = 7 ^ (1-x)

• math- precalculus -

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 -

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 -

1/6

• rwil ohsuycip -

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