Given: $C=10$, $C^a=3$, $C^b=5$, how to solve $C^{2a-b+1}$.

Question

Given: $C=10$, $C^a=3$, $C^b=5$, how to solve $C^{2a-b+1}$.

I would be very grateful if somebody show me how to solve this.

Thanks.

Answer 1

It’s simply a matter of using the laws of exponents:

$$\begin{align*} C^{2a-b+1}&=C^{2a}\cdot C^{-b}\cdot C^1\\\\ &=(C^a)^2\cdot\frac1{C^b}\cdot C\\\\ &=3^2\cdot\frac15\cdot 10\\\\ &=9\cdot 2\\\\ &=18\;. \end{align*}$$

Answer 2

It's like $C^a \cdot (C^a/C^b)\cdot C = 3 \cdot(3/5)\cdot 10 = 90/5 = 18$

I hope you understand

Answer 3

We are given that $C = 10, C^a = 3, C^b = 5$, and we want to evaluate $C^{2a - b + 1}$.

Then this becomes an exercise in rewriting. Note that $C^{2a - b + 1} = \dfrac{C \cdot (C^{a})^2}{C^b}$, and then you can substitute naively.