We are given the following two logarithmic expressions in the problem:

1. $\log_a b^2 = c$
2. $\log_b a = c - 1$

We are asked to express $a$ in terms of $b$.

Step 1: Simplify $\log_a b^2 = c$

Using the logarithmic property $\log_a (x^k) = k \cdot \log_a x$, we can simplify the first equation: $\log_a b^2 = 2 \log_a b$ So, we rewrite the first equation as: $2 \log_a b = c$ Thus: $\log_a b = \frac{c}{2}$ This gives us the relationship between $a$ and $b$ in terms of $c$.

Step 2: Simplify $\log_b a = c - 1$

The second equation is already in a simplified form, so we leave it as: $\log_b a = c - 1$

Step 3: Express $a$ in terms of $b$

To express $a$ in terms of $b$, we use the fact that $\log_b a = c - 1$ implies: $a = b^{c - 1}$

Step 4: Relating the two equations

Now, we use the value of $c$ from $\log_a b = \frac{c}{2}$ to help with this relationship. However, the second equation already expresses $a$ in terms of $b$, which is: $a = b^{c - 1}$

Hence, the expression for $a$ in terms of $b$ is: $a = b^{c - 1}$

Would you like to go into more details about this solution? Here are a few related questions for you:

1. Can you express $b$ in terms of $a$ from the given equations?
2. How would you solve if the logarithmic bases were different?
3. What other logarithmic identities could simplify the problem?
4. Can you solve the problem graphically using logarithmic functions?
5. How would the solution change if we were given different values of $c$?

Tip: Always try to simplify logarithmic equations using logarithmic identities like the power rule or change of base to avoid algebraic complexity.