Question

Solve:

$2{\log }_{10}x=1+{\log }_{10}(x+\frac{11}{10})$

Answer 1

We have,

$2{\log }_{10}x=1+{\log }_{10}(x+\frac{11}{10})$

We know that,

${\log }_{10}10=1$

So,

$2{\log }_{10}x={\log }_{10}10+{\log }_{10}(x+\frac{11}{10})$

$\Rightarrow {\log }_{10}{x}^2={\log }_{10}10\times (x+\frac{11}{10})$

On comparing that,

$x^2=10\times (x+\frac{11}{10})$

$\frac{x^2}{10}=\frac{10x+11}{10}$

Now,

$x^2=10x+11$

$\Rightarrow x^2-10x-11=0$

$\Rightarrow x^2-(11-1)x-11=0$

$\Rightarrow x^2-11x+x-11=0$

$\Rightarrow x(x-11)+1(x-11)=0$

$\Rightarrow (x-11)(x+1)=0$

$\Rightarrow x-11=0,x+1=0$

$\Rightarrow x=-1,11$

Hence, this is the answer.