How to solve $6^{2x}-10\cdot 6^x=-21$ using logarithms? 
What do I do with $\large 6^{2x}-10\cdot 6^x=-21$?

Since $6$ and $-60$ are not of the same base (nor can they be written as exponents of the same base cleanly) I am having trouble solving for $x$.

Let $6^x=y$
$$6^{2x}-10\cdot 6^x=-21$$
$$y^2-10y=-21$$
$$y^2-10y+21=0$$
Factor.
$$(y-3)(y-7)=0$$
$$y=3, \ 7$$
Replace $y$ with $6^x$
$$6^x = 3, \ 7$$
I will solve for both equations separately. Let's start with $6^x=3$
$$6^x=3$$
$$\ln(6^x)=\ln(3)$$
$$x\ln(6)=\ln(3)$$
$$x=\dfrac{\ln(3)}{\ln(6)}$$
Now for $6^x=7$
$$6^x=7$$
$$\ln(6^x)=\ln(7)$$
$$x\ln(6)=\ln(7)$$
$$x=\dfrac{\ln(7)}{\ln(6)}$$
The solutions are:
$$\displaystyle \boxed{x=\dfrac{\ln(3)}{\ln(6)}, \ \dfrac{\ln(7)}{\ln(6)}}$$
