Question

What is :(49^x)(7^x^2)=2401^2

Answer 1

$(49^x)(7 ^{x^2})=2401^2 \\ \\( (7^2)^x)(7 ^{x^2})= (7^4) ^2 \\ \\ 7 ^{2x}\cdot 7^{x^2}=7^{8} \\ \\ 7 ^{2x+x^2} =7^{8} \\ \\2x+x^2 =8$

 $x^2-2x-8 = 0\\ \\ a=1, \ b=-2, \ c=-8 \\ \\\Delta =b^2-4ac = (-2)^2 -4\cdot1\cdot (-8) = 4 +32 =36 \\ \\x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{2-\sqrt{36}}{2 }=\frac{ 2-6}{2}=\frac{-4}{2}=-2 \\ \\x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{2+\sqrt{36}}{2 }=\frac{ 2+6}{2}=\frac{8}{2}= 4 \\ \\Answer: \ x=-2 \ \ or \ \ x=4$

Answer 2

$(49^x)(7^{x^2})=2401^2\\ \\7^{2x} \cdot7^{x^2}=(7^4)^2\\ \\7^{2x+x^2}=7^8\ \ \ \Leftrightarrow\ \ \ 2x+x^2=8\ \ \ \Leftrightarrow\ \ \ x^2+2x-8=0\\ \\x^2-2x+4x-8=0\ \ \ \Leftrightarrow\ \ \ x(x-2)+4(x-2)=0\\ \\(x-2)(x+4)=0\ \ \ \Leftrightarrow\ \ \ ( x-2=0\ \ \ \vee\ \ \ x+4=0)\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-4$