I have a Long Power

Simplify the expression. $\begin{aligned} x^{2\log_x(x+3)}&=x^{\log_x(x+3)^2}\\ &=(x+3)^2=16 \end{aligned}$

If $(x+3)^2=16,$ then $x=-7$ or $x=1.$ So the answer is $2,$ right? No. If you try to resubstitute those values into the original equation, you would see that $1^{2\log_14}=16$ or $(-7)^{2\log_{-7}-4}=16,$ which is nonsense. The base of a logarithm cannot be $1$ or a negative number. Both $-7$ and $1$ are extraneous solutions, so the equation has $\boxed{0}$ roots.

My solution evolved in a different fashion :

$x^{2\log_x(x+3)}=\left(x^{\log_x(x+3)}\right)^2=16$

Remembering that $x>0$ (for the log to make sense)

$x^{\log_x(x+3)}=4$

Taking the natural logarithm and using $\log_x(x+3)=\frac{\ln(x+3)}{\ln(x)}$ ,

$\frac{\ln(x+3)}{\ln(x)}\ln(x)=\ln(4)$

If $\ln(x) \ne 0$ , or $x \ne 1$ , then

$\ln(x+3)=\ln(4) \implies (x+3)=4 \implies x=1$

But, this cannot be the solution as $x \ne 1$ . Hence, number of solutions is $\boxed{0}$

$x^{\log_{x}(x+3)}=y \Rightarrow y^2=16 \therefore y=\pm 4$

We know that $m^{\log_{m}{n}}=n \Rightarrow y=x+3$

$x+3=4 \Rightarrow x=1$ but any logartithm base can be $1$ .

$x+3=-4 \Rightarrow x=-7$ but any logartithm base can be negative.

$\therefore x^{2\log_{x}(x+3)}=16$ has $\boxed{0}$ roots.