Zadania w załączniku(logarytmy, potęgi)proszę o pomoc :)).... Question from @Kropqa5791

December 2018 1 7 Report

Zadania w załączniku
(logarytmy, potęgi)

proszę o pomoc :))

loitzl9006 10.
a) $\frac2{\sqrt[3]{32}}=\frac2{2^{\frac53}}=2^{1-\frac53}=2^{-\frac23}$

b)
$\sqrt{\frac12\cdot\sqrt{\frac12}}=\sqrt{2^{-1}\cdot\sqrt{2^{-1}}}=\sqrt{2^{-1}\cdot2^{-\frac12}}=\sqrt{2^{-1-\frac12}}=\sqrt{2^{-\frac32}}=\\=(2^{-\frac32})^{\frac12}=2^{-\frac34$

c)
$\frac{4\cdot32\cdot\frac1{\sqrt2}}{(2^3)^4:\sqrt[3]{64}}=\frac{2^2\cdot2^5\cdot2^{-\frac12}}{2^{12}:4}=\frac{2^{2+5-\frac12}}{2^{12}:2^2}=\frac{2^{\frac{13}2}}{2^{10}}=2^{\frac{13}2-10}=2^{-\frac72}$

11.
a)
$7^{2-\sqrt3}\cdot49^{1+\frac{\sqrt3}2}=7^{2-\sqrt3}\cdot(7^2)^{1+\frac{\sqrt3}2}=7^{2-\sqrt3}\cdot7^{2+\sqrt3}=7^{2-\sqrt3+2+\sqrt3}=\\=7^4=2401$

b)
$\log_a216=3 \ \to \ a^3=216 \ \to \ a^3=6^3 \ \to \ a=6$

c)
$\log_{\sqrt2}x=6 \ \to \ x=(\sqrt2)^6 \ \to \ x=(2^{\frac12})^6 \ \to \ x=2^3=8$

d)
$\log_a\sqrt3=\frac14 \ \to \ a^{\frac14}=\sqrt3 \ \to \ (a^{\frac14})^4=(\sqrt3)^4 \ \to \ a=(3^{\frac12})^4=3^2=9$

e)
$\log_42-\log_432-6\underbrace{\log_41}_0=\log_4(\frac2{32})-6\cdot0=\log_4(\frac1{16})-0=\\=\log_4(4^{-2})-0=-2-0=-2$

f)
$\log_312+2\log_36-4\log_32=\log_312+\log_36^2-\log_32^4=\\=\log_312+\log_336-\log_316=\log_3(\frac{12\cdot36}{16})=\log_327=3$

12.
a)
$\log_3x=2+\log_32 \\ \log_3x-\log_32=2 \\ \log_3(\frac x2)=2 \ \to \ \frac x2=3^2 \ \to \ \frac x2=9 \ \to \ x=18$

b)
$\log x=-1+2\log5 \\ \log x-2\log5=-1 \\ \log x-\log5^2=-1 \\ \log x-\log25=-1 \\ \log (\frac x{25})=-1 \ \to \ \frac x{25}=10^{-1} \ \to \ x=25\cdot10^{-1}=25\cdot\frac1{10}=2.5$

13.
$\overline{x}=\frac{\log_412-\log_43+2}3=\frac{\log_4(\frac{12}3)+2}3=\frac{\log_44+2}3=\frac{1+2}3=\frac33=1$

14.
$\frac{\log3+\log4}{\log24-\log2}=\frac{\log(3\cdot4)}{\log(\frac{24}2)}=\frac{\log12}{\log12}=1$

15.
$a=|\underbrace{\sqrt5-3}_{ujemna}|+|\underbrace{\sqrt5-2}_{dodatnia}|=-(\sqrt5-3)+(\sqrt5-2)=\\=-\sqrt5+3+\sqrt5-2=1$

16.
$\sqrt{(x-1)^2}=\sqrt{1\frac9{16}} \\ |x-1|=\sqrt{\frac{25}{16}} \\ |x-1|=\frac54 \\ x-1=\frac54 \ \vee \ x-1=-\frac54 \\ x=1+\frac54 \ \vee \ x=1-\frac54 \\ x=\frac94 \ \vee \ x=-\frac14$

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