Question

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 (1) 3/2 (2) root 15 (3) (root 6) /2 (4) 15

Answer 1

Given that,

$\longrightarrow2^{x+y}=10$

Multiply both sides by 2.

$\longrightarrow2^{x+y}\times2=10\times2$

$\longrightarrow2^{x+y+1}=20\quad\quad\dots(1)$

But given that,

$\longrightarrow2^{y+z}=20$

Comparing this with (1) we get,

$\longrightarrow x+y+1=y+z$

$\longrightarrow x+1=z\quad\quad\dots(2)$

Given that,

$\longrightarrow2^{z+x}=30$

From (2),

$\longrightarrow2^{2x+1}=30$

$\longrightarrow2^{2x}\times2=30$

$\longrightarrow(2^x)^2=15$

$\longrightarrow\underline{\underline{2^x=\sqrt{15}}}$

Hence (2) is the answer.

Answer 2

Let $\sf 2^x$ be $\sf a,$ $\sf 2^y$ be $\sf b,$ and $\sf 2^z$ be $\sf c.$

We're given,

$\sf 2^{(x + y)} = 10$

$\longrightarrow \sf 2^x * 2^y = 10$

$\longrightarrow \sf ab = 10\quad\quad\dots(1)$

$\sf{\\}$

Similarly,

$\sf 2^{(y + z)} = 20$

$\longrightarrow \sf 2^y * 2^z = 20$

$\longrightarrow \sf bc = 10\quad\quad\dots(2)$

$\sf{\\}$

And,

$\sf 2^{(z + x)} = 20$

$\longrightarrow \sf 2^z * 2^x = 20$

$\longrightarrow \sf ca = 10$

$\longrightarrow \sf c = \dfrac{30}{a}\quad\quad\dots(3)$

$\sf{\\}$

From (3) and (2),

$\sf bc = 20$

$\longrightarrow \sf b * \dfrac{30}{a} = 20$

$\longrightarrow \sf b = \dfrac{2}{3}a \quad\quad\dots(4)$

$\sf{\\}$

From (4) and (1),

$\sf ab = 10$

$\longrightarrow \sf \bigg( \dfrac{2}{3}a\bigg)a = 10$

$\longrightarrow \sf \dfrac{2}{3}a^2 = 10$

$\longrightarrow \sf a^2 = \dfrac{10 * 3}{2}$

$\longrightarrow \sf a^2 = 15$

$\longrightarrow \sf a = \sqrt{15}$

$\longrightarrow \sf 2^x = \sqrt{15}$

$\sf{\\}$

Hence, the answer is,

$\longrightarrow \underline{\underline{\sf{\green{2^x = \sqrt{15}}}}}$