Question

rezolvari ecuatii logaritmice : log din baza 5 de x +1 la patrat = -3

Answer 1

$log_{5}$(x+1)²=-3
(x+1)²=$5^{-3}$
x²+2x+1=1/5³
x²+2x+1=1/125
125x²+250x+125-1=0
125x²+250x+124=0  am folosit formula $log_{a}$x=b⇒x=$a^{b}$
Δ=b²-4ac
Δ=250²-4·125·124
Δ=62500-62000
Δ=500
√Δ=10√5
x1=(-b+√Δ)/2a=(-250+10√5)/250
x2=(-b-√Δ)/2a=(-250-10√5)/250

Answer 2

[tex]\it log_5(x+1)^2=-3 \Rightarrow (x+1)^2 =5^{-3} \Rightarrow (x+1)^2 = \dfrac{1} {5^3} \Rightarrow \\ \\ \\ \Rightarrow \sqrt{(x+1)^2} = \sqrt{\dfrac{1}{5^3}} \Rightarrow |x+1| = \dfrac{1}{5\sqrt5} \Rightarrow |x+1| = \dfrac{\sqrt5}{25} \Rightarrow [/tex]

[tex]\it \Rightarrow x+1 = \pm \dfrac{ \sqrt5 }{25} \Rightarrow x= \pm \dfrac{ \sqrt5 }{25} -1 \Rightarrow \begin{cases} \it x_1= -\dfrac{ \sqrt5 }{25} -1 \\ \\ \\ \it x_2= \dfrac{ \sqrt5 }{25} -1 \end{cases}[/tex]