How do you solve ${(5^{x})}^{x + 1} = 5^{2 x + 12}$ $(5^x)^(x+1)=5^(2x+12)$ ?

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Answer 1 · 2016-03-09T10:30:09

$x = - 3$ $x=-3$ or $x = 4$ $x=4$

As ${(a^{m})}^{n} = a^{m n}$ $(a^m)^n=a^(mn)$ , ${(5^{x})}^{x + 1} = 5^{x (x + 1)} = 5^{x^{2} + x}$ $(5^x)^(x+1)=5^(x(x+1))=5^(x^2+x)$

Hence, $5^{x^{2} + x} = 5^{2 x + 12}$ $5^(x^2+x)=5^(2x+12)$ or

$x^{2} + x = 2 x + 12$ $x^2+x=2x+12$ or $x^{2} + x - 2 x - 12 = 0$ $x^2+x-2x-12=0$ or

$x^{2} - x - 12 = 0$ $x^2-x-12=0$ or

$x^{2} - 4 x + 3 x - 12 = 0$ $x^2-4x+3x-12=0$ or

$x (x - 4) + 3 (x - 4) = 0$ $x(x-4)+3(x-4)=0$ or

$(x + 3) (x - 4) = 0$ $(x+3)(x-4)=0$ or

$x = - 3$ $x=-3$ or $x = 4$ $x=4$

How do you solve (5x)x+1=52x+12(5^x)^(x+1)=5^(2x+12)?