Need help with solving logarithmic equations 2 2 x − 2 x − 6 =...

Hint:
In each case, you have quadratics in $b^x$:
In the first, put $y=2^x$. Then you have $y^2-y-6=0$
For the second, put $y=3^x$. Then you have $y^2-5y+4=0$
Note: Both quadratic equations factor quite nicely.
$\begin{array}{rl}(1)(2^x{)}^2-2^x-6=0& \stackrel{y=2^x}{⟹}{y}^2-y-6=(y-3)(y+2)=0\\ \\ & ⟺y=2^x=3\text{ or }y=2^x=-2\end{array}$
We can see immediately that in both solutions, $2^x$ must necessarily be positive, since $2>0$. So we're left with
$\begin{array}{}\text{(1)}& 2^x=3⟹\ln (2^x)=\ln 3⟺x\ln 2=\ln 3⟺x={\displaystyle \frac{\ln 3}{\ln 2}}\end{array}$
$\begin{array}{rl}(2)(3^x{)}^2-5(3^x)+4=0& \stackrel{y=3^x}{⟹}{y}^2-5y+4=(y-1)(y-4)=0\\ \\ & ⟺y=3^x=1\text{ or }y=3^x=4\end{array}$
So we have
$\begin{array}{}\text{(1)}& 3^x=1⟹\ln (3^x)=\ln 1⟺x\ln 3=\stackrel{=0}{\overbrace{\ln 1}}=0⟺x=0\end{array}$
And we have
$\begin{array}{}\text{(2)}& 3^x=4⟹\ln (3^x)=\stackrel{=\ln (2^2)=2\ln 2}{\overbrace{\ln 4}}⟺x\ln 3=2\ln 2⟺x={\displaystyle \frac{2\ln 2}{\ln 3}}\end{array}$

a. Let $2^x=m⟹m^2-m-6=0$
Solve the quadratic equation to get:
$(m+2)(m-3)=0⟹m=-2,m=3$
But remember $2^x=m$ so:
$\therefore 2^x=-2,2^x=3$
$\ln$ both sides of the equation to get:
$\ln 2^x=\ln 3⟹x\ln 2=\ln 3⟹x=\frac{\ln 3}{\ln 2}$
$\ln 2^x=\ln -2$
You cannot take the $\ln$ of a negative number so that solution is not applicable.
$\therefore x=\frac{\ln 3}{\ln 2}$
b. Let $3^x=n⟹n^2-5n+4=0$
Solve the quadratic equation to get:
$(n-4)(n-1)=0⟹n=1,n=4$
But remember $3^x=n$ so:
$\therefore 3^x=1,3^x=4$
$\ln$ both sides of the equation to get:
$\ln 3^x=\ln 1⟹x\ln 3=ln1$
but
$\ln 1=0⟹x=0$
$\ln 3^x=\ln 4⟹x\ln 3=\ln 4⟹x=\frac{\ln 4}{\ln 3}$
$\therefore x=0,x=\frac{\ln 4}{\ln 3}$