Question

Exercitiul 2, va rog!

Answer 1

Răspuns:

(4,+∞)

Explicație pas cu pas:

$log_{2}x+log_{4}x>3,~DVA~x>0\\log_{2}x+log_{2^{2}}x>3,~log_{2}x+\frac{1}{2} log_{2}x>3,~log_{2}x+log_{2}x^{\frac{1}{2} }>3,~log_{2}(x^{1}*x^{\frac{1}{2}})>3,~log_{2}x^{\frac{3}{2}}>log_{2}2^{3},~x^{\frac{3}{2}}>2^{3},~(x^{\frac{1}{2} })^{3}>2^{3},~deci ~x^{\frac{1}{2}} >2,~|^2,~x>4$

x∈(0,+∞)∩(4,+∞)=(4,+∞)

Answer 2

Condiția de existență a inecuației este $\it x>0.$

$\it log_4x=\dfrac{log_2x}{log_24}=\dfrac{log_2x}{2}$

Inecuația devine:

$\it log_2x+\dfrac{log_2x}{2}>3|_{\cdot2} \Rightarrow 2log_2x+log_2x>3\cdot2 \Rightarrow 3log_2x>3\cdot2|_{:3}\Rightarrow \\ \\ \\ \Rightarrow log_2x>2 \Rightarrow x>2^2 \Rightarrow x>4 \Rightarrow x\in(4,\ \infty)$