Question

Sa se afle numaru natural: a) 3^x×3^x+1=27^3 b) (2^x)^2×4^x-2=8^4 c) 2^x×4^x8^x=64

Answer 1

$a)3^{x}\cdot3^{x+1}=27^{3}\\ \\3^{x}\cdot3^{x+1}=3^{x+x+1} \\ \\27=3^{3} \\ \\3^{x+x+1}=(3^{3})^{3} \\ \\x+x+1=3\cdot3 \\ \\2x+1=9 \\ \\2x=9-1 \\ \\2x=8 \\ \\x=8:2 \\ \\x=4$

$b)(2^{x})^{2}\cdot4^{x}-2=8^{4} \\ \\(2^{x})^{2}\cdot(2^{2})^{x}-2^{1}=(2^{3})^{4}$

$(2^{x})^{2}\cdot(2^{2})^{x}-2^{1}=(2^{3})^{4}\\ \\ (2^{x})^{2}=(2^{2})^{x}\\ \\ 2^{x}=t \\ \\ t^{4}-2=4098\\ \\t=4098^{\frac{1}{4}$

$2^{x} =4098^{ \frac{1}{4}$

$x=\frac{ln(4098)}{4ln2}$

$c)2^{x}\cdot4^{x}\cdot8^{x}=64\\ \\2^{x}\cdot4^{x}=8^{x}\\ \\8^{x}\cdot8^{x} =64\\ \\8^{x}\cdot8^{x} =8^{1}\cdot8^{1} \\ \\ 8^{x+x}=8^{1+1} \\ \\x+x=2 \\ \\2x=2 \\ \\x=1$