Question

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

Answer 1

`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`