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

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Answer 1 · 2018-06-06T15:41:37

$x = \frac{1}{2} or x = - \frac{5}{2}$ $x = 1/2 or x = -5/2$

$81^{x^{3} + 2 x^{2}} = 27^{\frac{5 x}{3}}$ $81^(x^3 + 2x^2) = 27^((5x)/3)$

Note: $3^{4} = 81 and 3^{3} = 27$ $3^4 = 81 and 3^3 = 27$

$3^{4 (x^{3} + 2 x^{2})} = 3^{3 (\frac{5 x}{3})}$ $3^(4(x^3 + 2x^2)) = 3^(3((5x)/3))$

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

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

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

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

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;

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

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

$2x = 1 or 2x = -5$

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