How do you solve #4^(3x+2)times32^(x-2)=8^(3x-4)#?

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Answer 1 · 2016-09-15T18:14:23

The soln. is #x=-3.#

#4^(3x+2)xx32^(x-2)=8^(3x-4)#

#rArr (2^2)^(3x+2)*(2^5)^(x-2)=(2^3)^(3x-4)#

#rArr 2^(6x+4)*2^(5x-10)=2^(9x-12).........[(a^m)^n=a^(mn)]#

#rArr 2^(6x+4+5x-10)=2^(9x-12)...............[a^m*a^n=a^(m+n)]#

#rArr 2^(11x-6)=2^(9x-12)..............[a^m=a^n rArr m=n]#

#rArr 11x-6=9x-12#

#rArr 11x-9x=6-12#

#rArr 2x=-6#

#rArr x=-3#

This root satisfy the eqn.

Hence, the soln. is #x=-3.#

Answer 2 · 2016-09-15T19:16:48

#x = - 3#

We have: #4^(3 x + 2) times 32^(x - 2) = 8^(3 x - 4)#

Let's express the numbers in terms of #2#:

#=> (2^(2))^(3 x + 2) times (2^(5))^(x - 2) = (2^(3))^(3 x - 4)#

Using the laws of exponents:

#=> 2^(6 x + 4) times 2^(5 x - 10) = 2^(9 x - 12)#

#=> 2^(6 x + 4 + 5 x - 10) = 2^(9 x - 12)#

#=> 2^(11 x - 6) = 2^(9 x - 12)#

#=> 11 x - 6 = 9 x - 12#

#=> 2 x = - 6#

#=> x = - 3#

Therefore, the solution to the equation is #x = - 3#.