Give the solution1) 2−x9x=x−2x32) |x2−9x2+1|≤13) 16x−22x+1=34) log3(6−x)log3(x)=2

Question

Aaliyah Phillips · Accepted Answer

1)&amp;nbsp;2−x9x=x−2x3x3(2−x)=9x(x−2)x3(2−x)−9x(x−2)=0x(2−x)(x2+9)=0x=0&amp;nbsp;or&amp;nbsp;2−x=0&amp;nbsp;or&amp;nbsp;x2+9=0x=0&amp;nbsp;or&amp;nbsp;x=2&amp;nbsp;is the solution2)&amp;nbsp;|x2−9x2+1|≤1−1≤x2−9x2+1≤1−1≤x2−9x2+1&amp;nbsp;or&amp;nbsp;x2−9x2+1≤1−(x2+1)≤x2−9−x2−1≤x2−99−1≤2x28≤2x24≤x2±2≤xThere is no solution3)&amp;nbsp;16x−22x+1=324x−22x2=322x22x−22x2=322x(22x−2)=3Let&amp;nbsp;22x=m, thenm(m−2)=3m2−2m−3=0m2−3m+m−3=0m(m−3)+1(m−3)=0(m+1)(m−3)=0m+1=0&amp;nbsp;or&amp;nbsp;m−3=0m=−1&amp;nbsp;or&amp;nbsp;m=3Substitute&amp;nbsp;m=22x22x=−1&amp;nbsp;or&amp;nbsp;22x=322x=−1, which isn&#039;t possible22x=3Take ln on both sidesln⁡(22x)=ln⁡(3)2xln⁡(2)=ln⁡(3)x=ln⁡(3)2ln⁡(2)

Wilson Rivas · Accepted Answer

I did the forthlog3(6−x)log3(x)=2log3(6−x)=2log3(x)ln⁡(6−x)ln⁡(3)=2ln⁡(x)ln⁡(3)ln⁡(6−x)=2ln⁡(x)ln⁡(6−x)=ln⁡(x)26−x=x2x2+x−6=0x2+3x−2x−6=0x(x+3)−2(x+3)=0(x−2)(x+3)=0x=2 or x=−3but x=−3 doesn&#039;t satisfy the equation, thus x=2 is the solution.

Give the solution 1) \(\displaystyle{\frac{{{2}-{x}}}{{{9}{x}}}}={\frac{{{x}-{2}}}{{{x}^{{3}}}}}\) 2) \(\displaystyle{\left|{\frac{{{x}^{{2}}-{9}}}{{{x}^{{2}}+{1}}}}\right|}\leq{1}\) 3)

Answered question

Answer & Explanation

New Questions in Algebra I