Question

The number of real solutions of $\frac{{log}_{10}5+{log}_{10}(x^2+1)}{{log}_{10}(x-2)}=2$

• A. 2
• B. 3
• C. 1
• D. 0

Answer 1

The correct option is A 2

$\frac{log5+log(x^2+1)}{log(x-2)}=2$

$log\left(5\right(x^2+1\left)\right)=2log(x-2)$

$log\left(5\right(x^2+1\left)\right)=log(x-2{)}^2$

$5(x^2+1)=(x-2{)}^2$

$5x^2+5=x^2-4x+4$

$4x^2+4x+1=0$

Discriminant=$(4{)}^2-4(4)=0$

$=(2x{)}^2+2(2x\left)\right(1)+(1{)}^2=0$

$(2x+1{)}^2=0$

$x=\frac{-1}{2},\frac{-1}{2}$

Number of real solutions are 2