Question

va rog sa ma ajutati la aceste ecuatii logaritmice! e urgent! dau 55 de puncte va rog

Answer 1

Răspuns:

Explicație pas cu pas:

C.E. x>0

log in baz 4 din x= (1/2) log in baz 2 din x

log in baz 8 din x= (1/3) log in baz 2 din x

e ecuatia devibne

(1+1/2+1/3) log in baz 2 dinx=11/6

log in baz 2 dinx=1

x=2∈domeniului, deci convine

2

C.E.

x>0

x≠1

x²+x-2>0⇔x∈(-∞;-2)∪ (1;∞)

ramane x>1

apoi rezolvi ecuatia

x²+x-3=0

x1,2=(-1±√13)/2

din care convine doar (-1+√13)/2

Answer 2

 

$\displaystyle\bf\\1)\\\\log_2x+log_4x+log_8x=\frac{11}{6}\\\\log_2x+log_{2^2}x+log_{2^3}x=\frac{11}{6}\\\\log_2x+\frac{1}{2}log_{2}x+\frac{1}{3}log_{2}x=\frac{11}{6}\\\\log_2x\left(1+\frac{1}{2}+\frac{1}{3}\right)=\frac{11}{6}\\\\\\log_2x\left(\frac{6}{6}+\frac{3}{6}+\frac{2}{6}\right)=\frac{11}{6}\\\\\\\frac{6+3+2}{6}log_2x=\frac{11}{6}\\\\\\\frac{11}{6}log_2x=\frac{11}{6}\\\\\\log_2x=\frac{~~\dfrac{11}{6}~~}{\dfrac{11}{6}}\\\\\\log_2x=1\\\\x=2^1\\\\\boxed{\bf x=2}$

.

2)

$\displaystyle\bf\\log_x(4)=log_x(x^2+x-2)\\\\Conditie:~~~x>0~si~x\neq1\\\\x^2+x-2=4\\\\x^2+x-2-4=0\\\\x^2+x-6=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x_{12}=\frac{-1\pm\sqrt{1^2-4\cdot1\cdot(-6)}}{2\cdot1}\\\\x_{12}=\frac{-1\pm\sqrt{1+24}}{2}\\\\x_{12}=\frac{-1\pm\sqrt{25}}{2}\\\\x_{12}=\frac{-1\pm5}{2}\\\\x_1=\frac{-1-5}{2}=\frac{-6}{2}=-3~~Solutie~4rrespinsa~deoarece~-3<0\\\\x_2=\frac{-1+5}{2}=\frac{4}{2}=2\\\\Solutie~unica:~~~\boxed{\bf x=2}$