How to Solve : $ A =\frac{1}{6}\left((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3\right) $

Question

$ A =\frac{1}{6}\left((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2(24))^3\right).$ Solve for $2^A.$

(no calculators or graphs are permitted)

The way I went about solving this problem was using the properties of base $2$. For example, $2^x = 8$ , $x$ is obviously $3$, and $2^y = 16$, $y = 4 = x+1$, so adding $1$ to the exponent will give twice the answer. Since the arguments are twice each other, I used this pattern:

$ A =\frac{1}{6}\left((x)^3-(x+1)^3-(x+2)^3+(x+3)^3\right)$

$= \frac{1}{6}\left((x^3-(x^3 + 3x^2 + 3x + 1)-(x^3 + 6x^2 + 12x + 8)+(x^3 + 9x^2 + 27x + 27)\right)$

Simplifying this we get $A = (12x + 18)/6 = 2x + 3.$

So now I solved for $2^{2x+3}$ using $2^{2x} =2^x * 2^x = 3^2 = 9.$ Now, since we add $3$ to the exponent, we have $2^{2x+3} = (2^3)(9) = 72.$

Would this be the correct answer, and if not, what would be a good way of going about it?

Answer 1

If it were to be asked in the exam, you could save some time in the simplification exploiting the fact that $a^3-b^3 = (a-b).(a^2+ab+b^2)$

Rearanging in the following way

$$\frac{1}{6}\left((x+3)^3 - (x+2)^3 - \left((x+1)^3 - x^3)\right)\right)$$

$$\frac{1}{6}\left((x+3)^2 + (x+2)^2+(x+3)(x+2) - \left((x+1)^2 + x^2+x(x+1))\right)\right)$$

$$\frac{1}{6}\left((x^2+6x+9+x^2+4x+4+x^2+5x+6) - \left((x^2+2x+1+x^2+x^2+x)\right)\right)$$

$$\frac{1}{6}\left((3x^2+15x+19) - \left((3x^2+3x+1)\right)\right)$$ $$\frac{1}{6}\left(12x+18\right)$$

Answer 2

A good choice of substitution might help cut through the clutter of logs and nested parentheses, giving a visually pleasing and symmetrical solution.

Let $\text{L}^mn=(\log_2 n)^m$ and put $u=L3=\log_23$.

Hence

$$\begin{align} \text{L}^3 3&=(L3)^3&=u^3\\ \text{L}^3 6&=(L3+L2)^3&=(u+1)^3\\ \text{L}^3 12&=(L2+L4)^3&=(u+2)^3\\ \text{L}^3 24&=(L2+L8)^3&=(u+3)^3\end{align}$$

$$\begin{align} A&=\frac16 ((\log_2(3))^3-(\log_2(6))^3-(\log_2(12))^3+(\log_2 (24))^3)\\ &=\frac16 (L^33-L^36-L^312+L^324)\\ &=\frac 16 \left[u^3-(u+1)^3-(u+2)^3+(u+3)^3\right]\\ &=\frac 16 \left[ 12u+18\right]\\ &=2u+3\\ &=2L3+L8\\ &=L72=\log_2 72\\ 2^A&=72\qquad \blacksquare \end{align}$$

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Previous answer below: $$\begin{align}\text{L}^3 3&=(\log_2 3)^3&&=u^3\\ \text{L}^3 6&=(\log_2 6)^3&=(\log_2 3+\log_2 2)^3&=(u+1)^3\\ \text{L}^3 12&=(\log_2 12)^3&=(\log_2 3+\log_2 4)^3&=(u+2)^3\\ \text{L}^3 24&=(\log_2 24)^3&=(\log_2 3+\log_2 8)^3&=(u+3)^3\end{align}$$

$$\begin{align} A&=\frac16 ((\log_23)^3-(\log_2 6)^3-(\log_2 12)^3+(\log_2 24)^3)\\ &=\frac 16 \left[u^3-(u+1)^3-(u+2)^3+(u+3)^3\right]\\ &=\frac 16 \left[ 12u+18\right]\\ &=2u+3\\ &=2\log_2 3+\log_28\\ &=\log_2 72\\ 2^A&=72\qquad \blacksquare \end{align}$$