Asked by Alicia
Determine the exact value of x.
a. 3^2x = 5(3^x) +36
b. (1/8)^x-3 = 2x16^2x+1
c. 3^x2 + 20 = (1/27)^3X
a. 3^2x = 5(3^x) +36
b. (1/8)^x-3 = 2x16^2x+1
c. 3^x2 + 20 = (1/27)^3X
Answers
Answered by
Reiny
a) 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 ?
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 ?
Answered by
Alicia
Thank 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.
I'm really stuck on c. The x was made capital by mistake, it's actually supposed to be lower case.
Answered by
Alicia
The equation is actually 3^x^2 + 20 = (1/27)^3x
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