Die Lösung der Gleichung a x = b a^{x\ }=\ b a x = b ist

x = log ⁡ a ( b ) x\ =\ \log_a\left(b\right) x = lo g a ( b )

x = log ⁡ b ( a ) x\ =\ \log_b\left(a\right) x = lo g b ( a )

x = b a x\ =\ \sqrt{b^a} x = b a <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

x = a b x\ =\ \sqrt{a^b} x = a b <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

lg ⁡ ( 12 ) + lg ⁡ ( x ) = \lg\left(12\right)\ +\ \lg\left(x\right)\ =\ l g ( 1 2 ) + l g ( x ) =

lg ⁡ ( 12 x ) \lg\left(12x\right) l g ( 1 2 x )

lg ⁡ ( 12 + x ) \lg\left(12\ +\ x\right) l g ( 1 2 + x )

lg ⁡ ( 12 − x ) \lg\left(12\ -\ x\right) l g ( 1 2 − x )

lg ⁡ ( 12 x ) \lg\left(\frac{\text{12}}{\text{x}}\right) l g ( x 12 )

ln ⁡ ( 8 ) + 1 = \ln\left(8\right)+\ 1= ln ( 8 ) + 1 =

ln ⁡ ( 8 e ) \ln\left(8e\right) ln ( 8 e )

ln ⁡ ( 9 ) \ln\left(\ 9\right) ln ( 9 )

ln ⁡ ( 8 + e ) \ln\left(8\ +\ e\right) ln ( 8 + e )

ln ⁡ ( 8 ) + ln ⁡ ( 1 ) \ln\left(8\right)\ +\ \ln\left(1\right) ln ( 8 ) + ln ( 1 )

lg ⁡ ( 7 a 5 ) = \lg\left(\frac{7}{\frac{a}{5}}\right)= l g ( 5 a 7 ) =

lg ⁡ ( 7 ) − lg ⁡ ( a ) + lg ⁡ ( 5 ) \lg\left(7\right)-\lg\left(a\right)+\lg\left(5\right) l g ( 7 ) − l g ( a ) + l g ( 5 )

lg ⁡ ( 7 ) − lg ⁡ ( a ) − lg ⁡ ( 5 ) \lg\left(7\right)-\lg\left(a\right)-\lg\left(5\right) l g ( 7 ) − l g ( a ) − l g ( 5 )

a 5 lg ⁡ ( 7 ) \frac{a}{5}\lg\left(7\right) 5 a l g ( 7 )

lg ⁡ ( 7 ) + lg ⁡ ( a ) + lg ⁡ ( 5 ) \lg\left(7\right)+\lg\left(a\right)+\lg\left(5\right) l g ( 7 ) + l g ( a ) + l g ( 5 )

lg ⁡ ( z x + y ) = \lg\left(z\sqrt{x+y}\right)= l g ( z x + y <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ) =

lg ⁡ ( z ) + 1 2 lg ⁡ ( x + y ) \lg\left(z\right)+\frac{1}{2}\lg\left(x+y\right) l g ( z ) + 2 1 l g ( x + y )

lg ⁡ ( z ) + 1 2 lg ⁡ ( x ) + 1 2 lg ⁡ ( y ) \lg\left(z\right)+\ \frac{1}{2}\lg\left(x\right)+\ \frac{1}{2}\lg\left(y\right) l g ( z ) + 2 1 l g ( x ) + 2 1 l g ( y )

lg ⁡ ( z ) + lg ⁡ ( x ) ⋅ lg ⁡ ( y ) \lg\left(z\right)+\lg\left(\sqrt{x}\right)\cdot\lg\left(\sqrt{y}\ \right) l g ( z ) + l g ( x <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ) ⋅ l g ( y <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> )

lg ⁡ ( z ) + lg ⁡ ( x 0 , 5 ⋅ y 0 , 5 ) \lg\left(z\right)+\lg\left(x^{0,5}\ \cdot y^{0,5}\right) l g ( z ) + l g ( x 0 , 5 ⋅ y 0 , 5 )

Logarithmusgesetze

Authored by Monika Ojekhile

Mathematics

5th Grade - University

Used 9+ times

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13 questions

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FILL IN THE BLANKS QUESTION

30 sec • 1 pt

Der Logarithmus von 27 zur Basis 3 ist

(a)

FILL IN THE BLANKS QUESTION

30 sec • 1 pt

Der Zehnerlogarithmus von 1000 ist

(a)

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

Die Lösung der Gleichung $a^{x\ }=\ b$ ist

$x\ =\ \log_a\left(b\right)$

$x\ =\ \log_b\left(a\right)$

$x\ =\ \sqrt{b^a}$

$x\ =\ \sqrt{a^b}$

FILL IN THE BLANKS QUESTION

30 sec • 1 pt

$\log_a\left(a^7\right)=?$

(a)

FILL IN THE BLANKS QUESTION

45 sec • 1 pt

$\log_x\left(36\right)=2$
Gib den Wert von x an

(a)

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Die Lösung der Gleichung lg(x) = 3 ist

x = 3

x = 10

x = 100

x = 1000

FILL IN THE BLANKS QUESTION

20 sec • 1 pt

Berechne den Logarithmus

$\log_a\left(1\right)$

(a)

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\lg\left(12\right)\ +\ \lg\left(x\right)\ =\$

$\lg\left(12\ +\ x\right)$

$\lg\left(12\ -\ x\right)$

$\lg\left(12x\right)$

$\lg\left(\frac{\text{12}}{\text{x}}\right)$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\ln(x)-2\ln(y)=$

$\ln(x\ -\ 2y)$

$\ln\left(x\ -\ y^2\right)^{ }$

$\ln\left(\frac{\text{x}}{\text{2y}}\right)$

$\ln\left(\frac{x}{y^2}\right)$

10.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

$\ln\left(8\right)+\ 1=$

$\ln\left(8e\right)$

$\ln\left(\ 9\right)$

$\ln\left(8\ +\ e\right)$

$\ln\left(8\right)\ +\ \ln\left(1\right)$

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Logarithmusgesetze

Der Logarithmus von 27 zur Basis 3 ist (a)

Der Zehnerlogarithmus von 1000 ist (a)

Die Lösung der Gleichung ax = ba^{x\ }=\ bax = b ist

log⁡a(a7)=?\log_a\left(a^7\right)=?loga​(a7)=? (a)

log⁡x(36)=2\log_x\left(36\right)=2logx​(36)=2 Gib den Wert von x an (a)

Die Lösung der Gleichung lg(x) = 3 ist

Berechne den Logarithmus log⁡a(1)\log_a\left(1\right)loga​(1) (a)

lg⁡(12) + lg⁡(x) = \lg\left(12\right)\ +\ \lg\left(x\right)\ =\ lg(12) + lg(x) =

ln⁡(x)−2ln⁡(y)=\ln(x)-2\ln(y)=ln(x)−2ln(y)=

ln⁡(8)+ 1=\ln\left(8\right)+\ 1=ln(8)+ 1=

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Der Logarithmus von 27 zur Basis 3 ist

(a)

Der Zehnerlogarithmus von 1000 ist

(a)

Die Lösung der Gleichung $a^{x\ }=\ b$ ist

$\log_a\left(a^7\right)=?$

(a)

$\log_x\left(36\right)=2$
Gib den Wert von x an

(a)

Berechne den Logarithmus

$\log_a\left(1\right)$

(a)

$\lg\left(12\right)\ +\ \lg\left(x\right)\ =\$

$\ln(x)-2\ln(y)=$

$\ln\left(8\right)+\ 1=$