algebra-precalculus
I am to find b when $\frac{36^{3b}}{36^{2b}}=216^{2-b}$. I got $b=\frac{12}{7}$ whereas my textbook says the solution is $\frac{6}{5}$
My working:
$$\frac{36^{3b}}{36^{2b}}=216^{2-b}$$
$$36^b=36^{6(2-b)}$$
$$b=6(2-b)$$
$$b=12-6b$$
$$7b=12$$
$$b=\frac{12}{7}$$
Where did I go wrong and how can I arrive at $\frac{6}{5}$?
The online textbook is here. It's exercise 9 at the end of that page. The solution is here under section exercises for chapter 6.6, exercise 9.
The $-2$ at the end is just translating the function down, it will not change if it is increasing or decreasing. Than you have a parabola, so you need to look at the vertex, which is at $-3$, since you have $(x-(-3))^2$. This is similar to $x^2$, just translated horizontally from $0$ to $-3$. You know that the parabola has values of $+\infty$ at $x=\pm\infty$, so it's decreasing from $-\infty$ to the vertex at $-3$, and increasing afterwards.
P.S. see @JWTanner 's comment
The root is $x = -4$. So you have to take $-4$ instead of $4$
Hence you have $2x^2-2x-3$
Best Answer
Note that $216 = 6^3$,
Hence we have $$6^{2b}=6^{3(2-b)}$$
$$2b=3(2-b)$$
$$5b=6$$