Question

Solve the equation
${10}^{(x+1)(3x+4)}-{2.10}^{(x+1)(x+2)}={10}^{1-x-x^2}$.

• A. $x=-1\pm \sqrt{\frac{1}{2}{\log }_{10}(1+\sqrt{11})}$
• B. $x=-1\pm \sqrt{\frac{1}{2}{\log }_{10}(1-\sqrt{11})}$
• C. $x=1\pm \sqrt{\frac{1}{2}{\log }_{10}(1+\sqrt{10})}$
• D. $x=1\pm \sqrt{\frac{1}{2}{\log }_{10}(1+\sqrt{11})}$

Answer 1

The correct option is A $x=-1\pm \sqrt{\frac{1}{2}{\log }_{10}(1+\sqrt{11})}$

the given equation is
${10}^{3x^2+7x+4}-{2.10}^{x^2+3x+2}=\frac{10}{{10}^{x+x^2}}$
$\Rightarrow$ ${10}^{4x^2+8x+4}-{2.10}^{2x^2+4x+2}=10$
$\Rightarrow$ ${10}^{4(x^2+2x+1)}-{2.10}^{2(x^2+2x+1)}=10$
$\Rightarrow$ ${10}^{4{(x+1)}^2}-{2.10}^{2{(x+1)}^2}=10$
$\Rightarrow$ ${\left\{{10}^{2{(x+2)}^2}\right\}}^2-2\left\{{10}^{2{(x+2)}^2}\right\}=10$ (i)
Let ${10}^{2{(x+1)}^2}=y$ (ii)
$\Rightarrow$ $y^2-2y=10$ $\Rightarrow$ $y^2-2y-10=0$
$\Rightarrow$ $y=\frac{2\pm \sqrt{4+40}}{2}$
$\Rightarrow$ $y=1\pm \sqrt{11}$
$\Rightarrow$ $y=1+\sqrt{11}$ (neglecting -ve sign as y> 0)
$\Rightarrow$ ${10}^{2{(x-1)}^2}=1+\sqrt{11}$
$\Rightarrow$ $2{(x+1)}^2={\log }_{10}(1+\sqrt{11})$
$\Rightarrow$ $2{(x+1)}^2={\log }_{10}(1+\sqrt{11})$
$\Rightarrow$ ${(x+1)}^2=\frac{1}{2}{\log }_{10}(1+\sqrt{11})$
$\Rightarrow$ $x+1=\pm \sqrt{\frac{1}{2}{\log }_{10}(1+\sqrt{11})}$
Hence, $x=-1\pm \sqrt{\frac{1}{2}{\log }_{10}(1+\sqrt{11})}$