We have, 9x2−18|x|+5=0 ⇒9|x|2−18|x|+5=0 ⇒(3|x|−1)(3|x|−5)=0 ⇒|x|=
1
3
or |x|=
5
3
⇒x=±
1
3
or x=±
5
3
Now, loge{(x+1)(x+2)} is defined, if (x+1)(x+2)>0 ⇒x<−2 or x>−1 Clearly, x=−
5
3
does not satisfy this condition. But all other values of x satisfy the above conditions. Hence, the number of solutions lying in the domain of definition of the given function is 3.