Question

If ${\log }_{x+1}(4x^3+9x^2+6x+1)+{\log }_{4x+1}(x^2+2x+1)=5,$ then the number of solution is

Answer 1

${\log }_{x+1}(4x^3+9x^2+6x+1)+{\log }_{4x+1}(x^2+2x+1)=5$
We know that,
$x+1>0,x+1\ne 1\Rightarrow x>-1,x\ne 04x+1>0,4x+1\ne 1\Rightarrow x>-\frac{1}{4},x\ne 0\therefore x\in (-\frac{1}{4},\infty )-\left\{0\right\}\cdots \left(1\right)$

Now,
${\log }_{x+1}(4x^3+9x^2+6x+1)+{\log }_{4x+1}(x^2+2x+1)=5\Rightarrow {\log }_{x+1}\left[\right(x+1\left)\right(4x^2+5x+1\left)\right]+{\log }_{4x+1}(x+1{)}^2=5\Rightarrow {\log }_{x+1}[(x+1{)}^2(4x+1\left)\right]+2{\log }_{4x+1}(x+1)=5\Rightarrow 2{\log }_{x+1}(x+1)+{\log }_{x+1}(4x+1)+2{\log }_{4x+1}(x+1)=5\Rightarrow {\log }_{x+1}(4x+1)+\frac{2}{{\log }_{x+1}(4x+1)}=3$

Assuming ${\log }_{x+1}(4x+1)=t$
$\Rightarrow t+\frac{2}{t}=3\Rightarrow t^2-3t+2=0\Rightarrow (t-1)(t-2)=0\Rightarrow t=1,2$

Now,
$t=1\Rightarrow {\log }_{x+1}(4x+1)=1$
$\Rightarrow 4x+1=x+1\Rightarrow x=0$
Which is not possible [from equation (1)]

$t=2\Rightarrow {\log }_{x+1}(4x+1)=2$
$\Rightarrow 4x+1=(x+1{)}^2\Rightarrow x^2-2x=0\Rightarrow x(x-2)=0\Rightarrow x=2(\because x\ne 0)$

Hence, the number of solution is $1$.