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December 17, 2014

December 17, 2014

Posted by **Alicia** on Sunday, January 16, 2011 at 12:42pm.

a. 3^2x = 5(3^x) +36

b. (1/8)^x-3 = 2x16^2x+1

c. 3^x2 + 20 = (1/27)^3X

- MATHHH -
**Reiny**, Sunday, January 16, 2011 at 4:10pma) let 3^x = y

so you have y^2 - 5y - 36=0

(y-9)(y+4) = 0

y = 9 or y = -4

3^x = 9

3^x = 3^2

x = 2

or 3^x = -4 , not possible

b) the trick here is to see that all bases are powers of 2

1/8 = 2^-3

16 = 2^4

so (1/8)^(x-3) = 2x16^(2x+1)

(2^-3)^(x-3) = 2(2^4)^(2x+1)

2^(-3x+9) = 2(2)^(8x+4)

2^(-3x+9) = 2^(8x+5)

then

-3x+9 = 8x+5

you can finish it ...

c) I see no easy way to do this one.

Why is the a capital X ?

- MATHHH -
**Alicia**, Monday, January 17, 2011 at 10:17amThank you for the help,

I'm really stuck on c. The x was made capital by mistake, it's actually supposed to be lower case.

- MATHHH -
**Alicia**, Monday, January 17, 2011 at 10:33amThe equation is actually 3^x^2 + 20 = (1/27)^3x

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