I bet it's fun

we have $\log_{2x+5}{(x^2-19)}=1$ .Now we know that if $\log_x y = z$ ,then $x^z = y$ .we use this in our problem to get: $2x+5=x^2-19 \\ \implies x^2-2x-24=0 \implies x^2-6x+4x-24=0 \\ \implies (x-6)(x+4)=0 \\ \implies x=6,-4.$ but $\log_{2x+5}{(x^2-19)}$ is undefined for negative value of $x$ . $\therefore x=6$

Please make the options better. 6 is the only number which lies in the domain of the given function

$\log_{2x+5} (x^2-19)=1$

$\text{Let } 2x+5=b, x^2-19=x,1=y$

$\text {When} \log_b x=y , b^y=x$

$\therefore (2x+5)^1=x^2-19$

$\therefore 2x+24=x^2$

$\therefore -x^2+2x+24=0$

$\therefore -x^2-4x+6x+24=0$

$\therefore -x (x+4)+6 (x+4)=0$

$\therefore (-x+6)(x+4)=0$

$\text{Hence, x=6 or x=-4}$

$\text{Substituting x=6 into the logarithm, }$

$\log_{2(6)+5} (6^2-19)=\log_{17} (17)=1$

$\text {Hence , x=6 (Proven)}$

Log -3/ Log -3 = 1 for x = -4 is also an answer in complex number realm as (-3)^1 = (-3).