Question

Exercitiu logaritmi...

Answer 1

$(\log_{10} x)^2+\log_{10}\sqrt{x}-\frac{3}{2}=0\\ \\ (\log_{10} x)^2+\log_{10}(x^{\frac{1}{2}})-\frac{3}{2}=0\\ \\(\log_{10} x)^2+\frac{1}{2}\cdot \log_{10}x-\frac{3}{2}=0\\ \\ Conditiile \ de \ existenta: \ x>0\\ \\Notam \ \log_{10}x=t\\ \\t^2+\frac{1}{2}\cdot t -\frac{3}{2}=0\\ \\ \Delta=(\frac{1}{2})^2-4\cdot 1 \cdot (-\frac{3}{2})\\ \\ \Delta=\frac{1}{4}+\frac{12}{2}=\frac{1}{4}+\frac{24}{4}=\frac{25}{4}\\ \\ t_{1, \ 2}=\frac{-\frac{1}{2}\pm \sqrt{\frac{25}{4}}}{2\cdot 1}=\frac{-\frac{1}{2}\pm \frac{5}{2}}{2}=\frac{\frac{-1\pm 5}{2}}{2}=\frac{-1 \pm 5}{4}\\ \\ t_1=\frac{-1+5}{4}=1\\ \\ t_2=\frac{-1-5}{4}=\frac{-3}{2} \\\\Cazul \ 1:\\ \\ \log_{10}x=1 \Rightarrow x=10^1=10>1\\ \\Cazul \ 2:\\ \\ \log_{10}x=\frac{-3}{2} \Rightarrow x=10^\frac{-3}{2}=\frac{1}{\sqrt{10^3}}=\frac{1}{10\sqrt{10}}\in (0; \ 1)\\ \\ \boxed{Raspuns: \ x=10}$

Answer 2

log²₁₀x+log₁₀√x-3/2=0

C.E.:x>0

      √x>0=>x>0

√x=x^(1/2)

log₁₀√x=log₁₀x^(1/2)=1/2*log₁₀x

fie log₁₀x=t

t²+1/2*t-3/2=0 |*2=>2t²+t-3=0

Δ=1+4*2*3=1+24=25

t₁,₂=(-1±5)/4

t₁=(-1+5)/4=4/4=1

t₂=(-1-5)/4=-6/4=-3/2

pt.t=1=>log₁₀x=1=>log₁₀ₓ=log₁₀10=>(bijectivitatea functiei logaritmice):x=10

pt.t=-3/2=>log₁₀x=-3/2=>x=10^(-3/2)=1/10^(3/2) ->subunitar =>N.C.