# How do you solve 49^x=7^(x^2-15)?

Nov 16, 2016

$x = - 3 \text{ OR } x = 5$

#### Explanation:

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${49}^{x} = {7}^{{x}^{2} - 15}$
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$\Rightarrow {\left({7}^{2}\right)}^{x} = {7}^{{x}^{2} - 15}$
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$\Rightarrow {7}^{2 x} = {7}^{{x}^{2} - 15}$
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Solving the equation of two powers having same base is
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determined by solving the equation formed from the equality
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of their powers.
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Therefore,
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$2 x = {x}^{2} - 15$
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$\Rightarrow - {x}^{2} + 2 x + 15 = 0$
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$\Rightarrow {x}^{2} - 2 x - 15 = 0 \text{ } E Q 1$
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Solving this equation is determined by Factorizing it.
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Factorization is determined by applying trial and error method:
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X^2 + SX + P = (X +a)(X + b)"
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$S = a + b \text{ and } P = a \times b$

In the equation :
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$\text{ S= -2" " and " " P = -15" " then a = -5 "and } b = + 3$
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Factorizing ${x}^{2} - 2 x - 15$ by using the explained method above:
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color(blue)(x^2-2x- 15 = (x+3)(x-5)" " EQ2
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Continuing to solve $E Q 1$
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${x}^{2} - 2 x - 15 = 0 \text{ } E Q 1$
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$\Rightarrow \left(x + 3\right) \left(x - 5\right) = 0 \text{ }$Substituting " color(blue)(EQ2)
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Therefore,
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$x + 3 = 0 \Rightarrow x = - 3$
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OR
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$x - 5 = 0 \Rightarrow x = 5$
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Hence,$\text{ " x = - 3" " Or " } x = 5$