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Răspuns :

1DianaMaria3

1)

a)log(3)81=log(3)3⁴=4log(3)(3)=4•1=4

b)log(5)✓125

[tex] log_{5}( \sqrt{5 {}^{3} } ) = log_{5}( {5}^{ \frac{2}{3} } ) = \\ = \frac{2}{3} log_{5}(5) = \frac{2}{3} [/tex]

[tex] \huge \green{ log_{x}(x) = 1}[/tex]

c)

[tex]lg \frac{1}{ \sqrt{100} } = lg \frac{1}{10 } = \\ = lg10 {}^{ - 1} = - 1lg10 = - 1[/tex]

[tex] \huge\green{lgx = log_{10}(x) }[/tex]

2)

Condiții de existenta:

___________

log(a)b

a≠1 a>0

b>0

__________

a)log2(2x-3)

2x-3>0

2x>3

x>3/2

x€(3/2;+infinit)

b)log✓3(x+5)

x+5>0

x>-5

x€(-5;+infinit)

Targoviste44

[tex]\it 1.\ a)\ log_3 81=4 \Leftrightarrow\ 3^4=81\\ \\ \\b)\ \ log_5\sqrt{125}=log_5 5^{\frac{3}{2}} =\dfrac{3}{2}log_5 5=\dfrac{3}{2}\\ \\ \\ c)\ lg\sqrt{\dfrac{1}{100}}=lg{\dfrac{1}{10}}=lg10^{-1}=-1\cdot lg10=-1[/tex]

[tex]\it 2.\ a)\ 2x-3>0 \Rightarrow 2x>3 \Rightarrow x>\dfrac{3}{2} \Rightarrow D=\Big(\dfrac{3}{2},\ \infty \Big)[/tex]

[tex]\it\ 3.\ a)\ log_6 24-log_6 4=log_6\dfrac{24}{4}=log_6 6=1[/tex]

[tex]\it 4.\ a)\ log_2 64-log_2 16 =log_2\dfrac{64}{16}=log_2 4 \\ \\ \\ Ecua\c{\it t}ia\ \ devine:\ \ log_2 x=log_2 4 \Rightarrow x=4[/tex]

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