Помогите решить
[tex]1) log_{2x+3} \frac{1}{4} +2=0[/tex]
[tex]2) log_{x} \sqrt{5} +log_{x} (5x)- \frac{9}{4} =(log_{x} \sqrt{5})^{2} [/tex]
[tex]1) log_{2x+3} \frac{1}{4} +2=0[/tex]
[tex]2) log_{x} \sqrt{5} +log_{x} (5x)- \frac{9}{4} =(log_{x} \sqrt{5})^{2} [/tex]
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Verified answer
1log(2x+3)1/4=-2
ОДЗ
{2x+3>0⇒2x>-3⇒x>-1,5
{2x+3≠1⇒2x≠-2⇒x≠-1
x∈(-1,5;-1) U (-1;∞)
(2x+3)^-2=1/4
2x+3=2
2x=-1
x=-0,5
2
ОДЗ
{x>0
{x≠1
x∈(0;1) U (1;∞)
4log(x)√5+4log(x)x+4log(x)5-9+4(log(x)√5)²
2log(x)5+4+4log(x)5-9=log²(x)5
log(x)5=a
a²-6a+5=0
a1+a2=6 U a1*a2=5
a1=1⇒log(x)5=1⇒x=5
a2=5⇒log(x)5=5⇒x=