If $\log_{x-1}(10-2x)=2$, find $x$.
I see that this is a logarithmic equation, so I want to rewrite it in exponential form to make it easier to work with. Using the definition of logarithms, I can write $\log_{x-1}(10-2x)=2$ as $(x-1)^2=10-2x$. Now I have a quadratic equation, which I can solve by expanding, simplifying, and factoring. Expanding the left side, I get $x^2-2x+1=10-2x$. Adding $2x$ to both sides, I get $x^2+1=10$. Subtracting $1$ from both sides, I get $x^2=9$. Taking the square root of both sides, I get $x=\pm 3$. However, I need to check if these values of $x$ make sense in the original equation, because sometimes logarithmic equations have extraneous solutions. If I plug in $x=3$, I get $\log_{3-1}(10-2\cdot 3)=2$, which simplifies to $\log_2(4)=2$, which is true. So $x=3$ is a valid solution. If I plug in $x=-3$, I get $\log_{-3-1}(10-2\cdot -3)=2$, which simplifies to $\log_{-4}(16)=2$. But this equation is not valid, because the base of a logarithm cannot be negative. So $x=-3$ is an extraneous solution. Therefore, the only solution to the original equation is $x=3$. 