+26387 2^x * 5^y = 100
2^y * 5^x = 1000
x+y = ?
\(\begin{array}{rcll} 2^x \cdot 5^y &=& 100 \\ 2^x \cdot 5^y &=& 10^2 \\\\ \hline \\ 2^y \cdot 5^x &=& 1000 \\ 2^y \cdot 5^x &=& 10^3 \\\\ \hline \\ (2^x \cdot 5^y)\cdot (2^y \cdot 5^x) &=& 10^2 \cdot 10^3 \\ 2^x \cdot 2^y \cdot 5^y \cdot 5^x &=& 10^{2+3} \\ 2^x \cdot 2^y \cdot 5^y \cdot 5^x &=& 10^{5} \\ 2^{x+y} \cdot 5^{y+x} &=& 10^{5} \\ (2\cdot 5)^{x+y} &=& 10^{5} \\ 10^{x+y} &=& 10^{5} \\\\ \mathbf{x+y} & \mathbf{=} & \mathbf{ 5 } \\ \end{array} \)
2^x * 5^y, = 100 2^y 5^x = 1000 x+y=?
Using Newton-Raphson method:
x= (2 log(2) - 3 log(5))/(log(2) - log(5)) ≈ 3.75647
y= (3 log(2) - 2 log(5))/(log(2) - log(5)) ≈1.24353
x + y = 3.75647 + 1.24353 =5
+9665 Another approach ---- take log each side.
\(2^x\cdot 5^y = 100 \rightarrow\;\;x\log 2 + y\log 5 = 2\)
\(2^y\cdot 5^x = 1000\rightarrow \;\; y\log 2 + x\log 5 = 3\)
After this we may solve it like a linear equation!!
\(\text{Add the 2 equations up:}\\ x\log 2 + y\log 2 + x\log 5 + y\log 5 = 5\\ (x+y)(\log 2 + \log 5) = 5 \leftarrow\text{Factorization}\\ (x+y)(\log 10) = 5 \leftarrow \text{Property of logarithms}\\ x+y = 5\)
