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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#