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49^x= 7^(x^2-15)49x=7x2−15
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rArr (7^2)^x = 7^(x^2-15)⇒(72)x=7x2−15
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rArr 7^(2x) = 7^(x^2-15)⇒72x=7x2−15
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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^2-152x=x2−15
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rArr -x^2+2x+15 = 0⇒−x2+2x+15=0
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rArrx^2-2x- 15 = 0" " EQ1⇒x2−2x−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)" X2+SX+P=(X+a)(X+b)
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S= a + b " and " P = axxbS=a+b and P=a×b
In the equation :
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" S= -2" " and " " P = -15" " then a = -5 "and " b=+3 S= -2 and P = -15 then a = -5 andb=+3
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Factorizing x^2-2x- 15x2−2x−15 by using the explained method above:
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color(blue)(x^2-2x- 15 = (x+3)(x-5)" " EQ2x2−2x−15=(x+3)(x−5) EQ2
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Continuing to solve EQ1EQ1
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x^2-2x- 15 = 0" " EQ1x2−2x−15=0 EQ1
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rArr(x+3)(x-5) = 0 " " ⇒(x+3)(x−5)=0 Substituting " color(blue)(EQ2)EQ2
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Therefore,
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x+3 = 0 rArr x =-3x+3=0⇒x=−3
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OR
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x - 5 = 0 rArr x = 5x−5=0⇒x=5
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Hence," " x = - 3" " Or " " x=5 x=−3 Or x=5
