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Solve the exponential equation $3^{3x^2+6x+3} = 1$
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RHS $=3^0$ by converting to same base, then equate exponents since bases are the same

$3x^2 + 6x + 3=0$

$x^2 + 2x + 1= 0$ use quadratic formula

$x=0.41$ or $x= - 2.41$
by Diamond (80.1k points)
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$3^(3x^2 + 6x +3) = 1$

Solve for $x:$

$3x^2 + 6x + 3 = \log_{3} (2)$

$3(x^2 + 2x + 1) = log_{3} (2)$

$3(x +1)^2 = log_{3} (2)$

$(x + 1)^2 = \frac{1}{3} \log_{3} (2)$

$x +1 = +$ or $-$ the square root of $\frac{3\log_{3} (2) }{3}$

$x_{1} = - 1.46$

$x_{2} = -0.54$
by Wooden (1.5k points)

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