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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.


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closed as off-topic by ᴡᴏʀᴅs, Najib Idrissi, N. F. Taussig, Umberto P., Ali Caglayan Oct 20 '15 at 21:08

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

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

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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.

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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

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this answer would be useful if you used latex – akkkk May 28 '12 at 10:03

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