Question

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

Answer 1

`x = -3" OR "x = 5`

Explanation:

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`49^x= 7^(x^2-15)`
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`rArr (7^2)^x = 7^(x^2-15)`
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`rArr 7^(2x) = 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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`2x = x^2-15`
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`rArr -x^2+2x+15 = 0`
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`rArrx^2-2x- 15 = 0" " EQ1`
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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 " and " P = axxb`

In the equation :
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`" S= -2" " and " " P = -15" " then a = -5 "and " b=+3`
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Factorizing `x^2-2x- 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 `EQ1`
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`x^2-2x- 15 = 0" " EQ1`
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`rArr(x+3)(x-5) = 0 " "`Substituting `" color(blue)(EQ2)`
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Therefore,
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`x+3 = 0 rArr x =-3`
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OR
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`x - 5 = 0 rArr x = 5`
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Hence,`" " x = - 3" " Or " " x=5`