Question 60
Let $$2^{x + y} = 10, 2^{y + z} = 20$$ and $$2^{z + x} = 30$$ where x, y and z are any three real numbers. The value of $$2^x$$ is
Solution
$$2^{x + y} = 10$$ ..... (1)
$$2^{y + z} = 20$$ ....... (2)
$$2^{z + x} = 30$$........ (3)
From (1), (2) and (3):
$$2^{2(x + y + z)} = 10 * 20 * 30 = 6000 = 400 * 15$$
$$2^{x + y + z} = 20\sqrt{15}$$ .... (4)
From (3) and (4):
$$\frac{2^{x + y + z}}{2^{y + z}} = \frac{20\sqrt{15}}{20}$$
$$2^x = \sqrt{15}$$
