Question

The real solutions of the equation $2^{x+2}{.5}^{6-x}={10}^{x^2}$ is/are

• A. $1$
• B. $2$
• C. $-{\log }_{10}\left(250\right)$
• D. $-{\log }_{10}4-3$

Answer 1

Answer: (B), (C)

The correct options are
B $-{\log }_{10}\left(250\right)$
C $2$

$2^{x+2}{.5}^{6-x}={10}^{x^2}$

$2^{x+2}{.5}^{6-x}=2^{x^2}{5}^{x^2}$

$(x+2){\log }_{10}2+(6-x){\log }_{10}5=x^2{\log }_{10}2+x^2{\log }_{10}5$

On comparing we get,

$x^2=x+2$ and $x^2=(6-x)$

$\Rightarrow (x-2)(x+1)=0$ and $(x+3)(x-2)=0$

$\Rightarrow x=2$

So, $x=2$ is one solution of the equation.

Writing the equation in quadratic form

$({\log }_{10}2+{\log }_{10}5)x^2+({\log }_{10}5-{\log }_{10}2)x-6{\log }_{10}5-2{\log }_{10}2=0$

Let the other root of equation be $r$.

$\Rightarrow 2\times r=\frac{-(6{\log }_{10}5+2{\log }_{10}2)}{{\log }_{10}2+{\log }_{10}5}$

$2\times r=-\frac{{\log }_{10}{5}^6\times 2^2}{{\log }_{10}2\times 5}$

$\Rightarrow r=-{\log }_{10}{(5^6\times 2^2)}^{1/2}$

$\Rightarrow r=-{\log }_{10}250$