Given exponential equations,a) 3^{(4x - }=11b) 6^{x + 3}=3^{x}To solve for x

Sub part a)
We have,
$3^{(ax-1)}=11$
Applying log to the base 10 on both sides,
$\Rightarrow {\log }_{10}{3}^{(4x-1)}={\log }_{10}11$
$\Rightarrow (4x-1){\log }_{10}3={\log }_{10}11$
$\Rightarrow 4x-1=({\log }_{10}11)/({\log }_{10}3)$
$\Rightarrow 4x-1={\log }_{3}11$
$\Rightarrow x=(1/4)[({\log }_{3}11)+1]$
$\Rightarrow x=0.795664585$
(Rounded to 9 decimals)
Identities used:
$({\log }_{a}x)/({\log }_{a}y)={\log }_{y}x$
$\log a^b=b\log a$
Sub part b)
We have,
$6^{x+3}=3^x$
Applying log to base 2 on both sides,
$\Rightarrow (x+3){\log }_{2}6=x({\log }_{2}3)$
$\Rightarrow (x+3){\log }_2(3\times 2)=x({\log }_{2}3)$
$\Rightarrow (x+3)[({\log }_{2}3)+({\log }_{2}2)]=x({\log }_{2}3)$
$\Rightarrow (x+3)[({\log }_{2}3)+1]=x({\log }_{2}3)$
$\Rightarrow x({\log }_{2}3)+3({\log }_{2}3)+x+3=x({\log }_{2}3)$
$\Rightarrow {\log }_2{3}^3+x+3=0$
$\Rightarrow {\log }_{2}9+x+3=0$
$\Rightarrow x=-3-{\log }_{2}9$
$\Rightarrow x=-6.169925001$
(Rounded to 9 decimals)
Identities used:
${\log }_{a}xy=({\log }_{a}x)+({\log }_{a}y)$
$\log a^b=b\log a$