What is the value of x in 81^(x^3 + 2x^2) = 27^((5*x)/3)?

1 Answer
Jun 6, 2018

Answer:

#x = 1/2 or x = -5/2#

Explanation:

#81^(x^3 + 2x^2) = 27^((5x)/3)#

Note: #3^4 = 81 and 3^3 = 27#

#3^(4(x^3 + 2x^2)) = 3^(3((5x)/3))#

#cancel3^(4(x^3 + 2x^2)) = cancel3^(3((5x)/3))#

#4(x^3 + 2x^2) = 3((5x)/3)#

#4(x^3 + 2x^2) = cancel3((5x)/cancel3)#

#4x^3 + 8x^2 = 5x#

Dividing through by #x#

#(4x^3)/x + (8x^2)/x = (5x)/x#

#(4x^(cancel3^2))/cancelx + (8x^(cancel2^1))/cancelx = (5cancelx)/cancelx#

#4x^2 + 8x = 5#

#4x^2 + 8x - 5 = 0#

Using Factorisation Method..

#2 and 10-> "factors"#

Proof: #10x - 2x = 8x and 10 xx -2 = -20#

Therefore;

#4x^2- 2x + 10x - 5 = 0#

Grouping the factors;

#(4x^2- 2x) + (10x - 5) = 0#

Factorising;

#2x(2x - 1) + 5(2x - 1) = 0#

Seperating the factors;

#(2x - 1) (2x + 5) = 0#

#2x - 1 = 0 or 2x + 5 = 0#

#2x = 1 or 2x = -5#

#x = 1/2 or x = -5/2#