SOLUTION: How is this done? 1.)log4 (3z) + log4x 2.)9^(x^2)=3^3x+2 3.)9^2x multiplied by (1/3)^x+2 =27 multiplied by (3^x)^-2

Question 733267: How is this done?
1.)log4 (3z) + log4x
2.)9^(x^2)=3^3x+2
3.)9^2x multiplied by (1/3)^x+2 =27 multiplied by (3^x)^-2
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1.)log4 (3z) + log4(x)
--
= log4[3z*x]
----
= log4(3xz)
------------------------------------
2.)9^(x^2)=3^3x+2
---
(3^2)^x^2 = 3^(3x+2)
---
3^(2x^2) = 3^(3x+2)
-----
2x^2 = 3x+2
2x^2-3x-2 = 0
Factor:
2x^2-4x+x-2 = 0
2x(x-2)+(x-2) = 0
(x-2)(2x+1) = 0
x = 2 or x = -1/2
Check for extraneous solutions.
======================================
3.)9^2x multiplied by (1/3)^x+2 =27 multiplied by (3^x)^-2
---
3^(4x)* 3^(-x-2) = 3^3*3^(-2x)
---
= 3^[4x-x-2] = 3^(3-2x)
---
3x-2 = 3x-2
-----
True for all values of "x".
=======================
Cheers,
Stan H.
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