Question

Puteti sa ma ajutati cu tema ex 7 pentru ca nam inteles ce trebuia sa fac

Answer 1

Răspuns:

Explicație pas cu pas:

a)

$(\frac{2}{3})^{n} =\frac{16}{81}=>(\frac{2}{3})^{n} =\frac{4^2}{9^2}=>(\frac{2}{3})^{n} =(\frac{4}{9})^2=>(\frac{2}{3})^{n}=(\frac{2^2}{3^2})^2=>(\frac{2}{3})^{n}= ((\frac{2}{3})^2)^2=>(\frac{2}{3})^{n}=(\frac{2}{3})^(2*2)=>(\frac{2}{3})^{n}=(\frac{2}{3})^4=>n=4$

b)

$(1\frac{1}{2})^n=3\frac{375}{1000}$

$(1\frac{1}{2})^n=(\frac{1*2+1}{2})^n=(\frac{3}{2})^n$ 1)

$3\frac{375}{1000}=3\frac{5*5*5*3}{5*2*5*2*5*2}=\frac{3*5*5*5*2*2*2+5*5*5*3}{5*2*5*2*5*2} =\frac{5*5*5(3*2*2*2+3)}{5*5*5*2*2*2}=\frac{3*2*2*2+3}{2*2*2}=\frac{3(2*2*2+1)}{2^3}=\frac{3*9}{2^3}=\frac{3*3^2}{2^3}=\frac{3^3}{2^3}=(\frac{3}{2})^3$

2)

Din 1) si 2) => $(\frac{3}{2})^n=(\frac{3}{2})^3=> n=3$

c)

$(\frac{1}{3^2})^n=\frac{1}{729}$

$(\frac{1}{3^2})^n=(\frac{1^2}{3^2})^n=((\frac{1}{3})^2)^n=(\frac{1}{3})^(2n)$  1)

$\frac{1}{729}=\frac{1}{9*81}=\frac{1}{9*9*9}=\frac{1^3}{9^3}=(\frac{1}{9})^3=(\frac{1^2}{3^2})^3=((\frac{1}{3})^2)^3=(\frac{1}{3})(2*3)=(\frac{1}{3})^6$   2)

Din 1) si 2) =>$(\frac{1}{3})^{2n}=(\frac{1}{3})^6=> 2n=6=> n=3$

d)

$(\frac{1}{7})^n=\frac{1}{343}=>(\frac{1}{7})^n=\frac{1}{7*49}=>(\frac{1}{7})^n=\frac{1}{7*7*7}=>(\frac{1}{7})^n=\frac{1^3}{7^3}=>(\frac{1}{7})^n=(\frac{1}{7})^3=>n=3$

e)$(2\frac{1}{2})^n=\frac{125}{8}$

$(2\frac{1}{2})^n=(\frac{2*2+1}{2})^n=(\frac{5}{2})^n$  1)

$\frac{125}{8}=\frac{5*25}{2*4}=\frac{5*5*5}{2*2*2}=\frac{5^3}{2^3}=(\frac{5}{2})^3$   2)

Din 1) si 2) => $(\frac{5}{2})^n=(\frac{5}{2})^3=>n=3$

f)

$(\frac{1}{8})^n=\frac{1}{4^3}=>(\frac{1}{2*4})^n=\frac{1^3}{4^3}=>(\frac{1}{2*2*2})^n=(\frac{1}{4})^3=>(\frac{1^3}{2^3})^n=(\frac{1^2}{2^2})^3=>((\frac{1}{2})^3)^n= ((\frac{1}{2})^2)^3=>(\frac{1}{2})^(3n)=(\frac{1}{2})^(6)=>3n=6=> n=2$