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Schule/Mathe/Schuljahr 2/Log Wachstum/Aufgaben Mathe.md

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+[[Sätze Mathe 10]]
+![[Pasted image 20260310105957.png]]
+$10^3$
+$10^{-2}$
+$2^4$
+$\frac{1}{2}^{-2}$
+$5^4=625$
+![[Pasted image 20260310110330.png]]
+1) $2$
+2) $3$
+3) $-2$ Frage: $2^{-2}=81$
+4) $1$ Frage: $7^1=7$
+5) $6$ Frage: $10^6=1'000'000$
+![[Pasted image 20260310110851.png]]
+a) $log_7(5)=x$
+b) $log_p(q)=x$
+c) $log_a(\frac{p}{q})=x$
+d) $log_\frac{a}{b}(c)=x$
+![[Pasted image 20260310111202.png]]
+a) $x=4$
+b) $x=2$
+c) $x=\frac{1}{2}$
+d) $x=$
+![[Pasted image 20260310112501.png]]
+a) $3*log(b)+5*log(d)$
+b) $log(12)+log(b)+n*log(d)-log(5)-log(c)-log(c)-r*log(f)$
+c) $log(5)+4*log(c)-log(8)-6*log(d)$
+d) $5*log(x-4)$
+![[Pasted image 20260310113318.png]]
+a) $log(1)$
+b) $log(1)$
+c) $log(1)$
+d) $log(150)$
+![[Pasted image 20260310114734.png]]
+a) F
+b) 
+c) 
+d) 
+e) 
+f) 
+![[Pasted image 20260317110316.png]]
+a) $$log_2(b)-log_2(c)-log_2(d)$$
+b)$$2+\frac{1}{2}*log_c(a)-\frac{1}{3}log_c(b)$$
+![[Pasted image 20260317111054.png]]
+a)$$log(\frac{32}{27})$$
+b)$$log(\frac{2*a^3\sqrt{n}}{m^4})$$
+![[Pasted image 20260317112016.png]]
+a)$$2^{10}=1024$$
+b)$$\frac{1}{1000}$$
+d)$$a=x$$
+![[Pasted image 20260317112724.png]]
+a) 
+b) 
+c) 
+![[Pasted image 20260317115307.png]]
+a)$$log_2(11)=x=3.459$$
+b)$$log_e(\pi)=x$$
+c)$$log_{0.8}(0.005)=x=23.74$$
+![[Pasted image 20260317120251.png]]
+a)$$$$
+b)$$log(3^{2x})=log(4*5^{x+3})$$
+$2x*log(3)=log(4)+x*log(5)+3*log(5)$ ->$2x*log(3)-x*log(5)=log(4)+3*log(5)$-> $x*(2*log(3)-log(5))=log(4)+3*log(5)$ -> $x=\frac{log(4)+3*log(5)}{2*log(3)-log(5)}$ Halt in Tr dann und dann lösung

Schule/Mathe/Schuljahr 2/Log Wachstum/Sätze.md

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+[[Sätze Mathe 10]]
+Der 2er-Logarithmus des Produkts $4*128$ entspricht also gerade der Summe der 2er-Logarithmen von
+