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December 20, 2014

December 20, 2014

Posted by **kayla** on Monday, December 1, 2008 at 3:50pm.

3^(x^3)=9^x

2^x=10

(4^x)-(2^x)=0

- precalculus -
**Reiny**, Monday, December 1, 2008 at 4:21pmfor the first you don't even need logs

3^(x^3)=9^x

3^(x^3)=(3^2)^x

3^(x^3) = 3^(2x)

so x^3 = 2x

x^3 - 2x = 0

x(x^2 - 2) = 0

x = 0 or x = ±√2

for second, take logs of both sides

log(2^x) = log 10

xlog2 = 1

x = 1/log2

for the last

4^x = 2^x

2^(2x) = 2^x

2x = x

x = 0

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