If $\log_{b/a} a^2 b = 2$, where $a$ and $b$ are positive real numbers, then find $\log_{ab} b$.
We can rewrite $\log_{b/a} a^2 b = 2$ as $(b/a)^2a^2b=(b/a)^2\cdot ab = b^2$. Therefore, $\log_ab^2 = \boxed{2}$.
To solve this question, we can use the properties of logarithms.
We have $\log_{b/a} (a^2 b) = 2$.
First, we can rewrite the logarithm expression using the change of base formula:
$$\frac{\log_{10}(a^2b)}{\log_{10}(b/a)} = 2$$
Next, we simplify the expression inside the logarithm:
$$\frac{\log_{10}(a^2b)}{\log_{10}(b)-\log_{10}(a)} = 2$$
Expanding the numerator using the product rule for logarithms:
$$\frac{\log_{10}a^2 + \log_{10}b}{\log_{10}b-\log_{10}a} = 2$$
Using the power rule for logarithms:
$$\frac{2\log_{10}a + \log_{10}b}{\log_{10}b-\log_{10}a} = 2$$
Now, multiplying both sides by $\log_{10}b - \log_{10}a$ to clear the denominator:
$$2(2\log_{10}a + \log_{10}b) = (\log_{10}b - \log_{10}a)(2)$$
Simplifying:
$$4\log_{10}a + 2\log_{10}b = 2\log_{10}b - 2\log_{10}
Moving all terms involving logarithms to one side:
$$4\log_{10}a + 2\log_{10}b + 2\log_{10}a - 2\log_{10}b = 0$$
Combining like terms:
$$6\log_{10}a = 0$$
Finally, dividing both sides by 6:
$$\log_{10}a = 0$$
Therefore, $\log_{10}a = 0$.
Since $\log_{ab} b = \frac{\log_{10}b}{\log_{10}ab}$, we can substitute $\log_{10}a = 0$ into the expression:
$$\log_{ab}b = \frac{\log_{10}b}{\log_{10}ab} = \frac{\log_{10}b}{\log_{10}a + \log_{10}b} = \frac{\log_{10}b}{0 + \log_{10}b} = \frac{\log_{10}b}{\log_{10}b} = 1$$
Therefore, $\log_{ab}b = 1$.