Na szybko 40 punktów plis.... Question from @kajetantoton23p4x3oh

andrzejsz2

Odpowiedź:

Szczegółowe wyjaśnienie:

a) log3(3)^20 = 20

b) log5(5)^0,1 = 0,1

c) log0,3(√0,3) = log0,3(0,3)^(1/2) = 1/2

d) log√3(√3^10) = 10log√3(√3) = 10

e) 4^[log4(5)] = 5

f) 3^[7log3(2)] = 3^[log3(2)^7 = 3^[log3(128)] = 128

g) 25^[log5(3)] = 5^[2log5(3)] = 5^[log5(9)] = 9

h) 7^[log49(4)] = 7^[log7(4)/log7(49)] = 7^[log7(4)/2] = 7^[1/2 log7(4)] =

7^[log7(2)] = 2

a) log3(2/9) + log3(1/6) = log3(2/54) = log3(1/27) = log3(3)^(-3) = -3

b) log2(10) + log2(1,6) = log2(16) = 4

c) log (25) + log (4) = log (100) = 2

d) log5(√10) - log5(√2) = log5(√5) = 1/2

e) log (9) - log (0,9) = log 10 = 1

f) log8(7) - log8(14) = log8(1/2) = x czyli:

8^x = 1/2

2^(3x) = 2^(-1)

3x = -1

x = -1/3

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jotka12

Odpowiedź:

zad2

a)log₃3²⁰=20log₃3=20

b) $log_{5}}5^{0,1}}=0,1log_{5}}5=0,1$

c) $log_{0,3}}\sqrt{0,3} =log_{0,3}}0,3^{\frac{1}{2} }=\frac{1}{2} log_{0,3}}0,3=\frac{1}{2}$

d) $log_{\sqrt{3} }\sqrt{3^{10}} =10log_{\sqrt{3} }\sqrt{3} =10$

e) $4^{log_{4}5=5$

f) $3^{7log_{3}2=3^{log_{3}2^7=2^7=128$

g) $25^{log_{5}3}=5^{2log_{5}3}=5^{log_{5}3^2}=3^2=9$

h) $7^{log_{49}4}}=7^{log_{7^2}4}=7^{\frac{1}{2} log_{7}4}=7^{log_{7}4^{\frac{1}{2}} =4^{\frac{1}{2} }=\sqrt{4} =2$

zad3

a) $log_{3}\frac{2}{9} +log_{3}\frac{1}{6} =log_{3}(\frac{2}{9} *\frac{1}{6} )\=log_{3}\frac{1}{27} =log_{3}3^{-3}=-3$

b) $log_{2}10+log_{2}1,6=log_{2}(10*1,6})=log_{2}16=log_{2}2^4=4$

c)log25+log4=lof(25*4)=log100=log10²=2

d) $log_{5}\sqrt{10} -log_{5}\sqrt{2} =log_{5}\frac{\sqrt{10} }{\sqrt{2} } }=log_{5}\sqrt{5} =log_{5}5^{\frac{1}{2}}=\frac{1}{2}$

e) $log9-log0,9=log\frac{9}{0,9} =log10=1$

f) $log_{8}7-log_{8}14=log_{8}\frac{7}{14} =log_{2^3}\frac{1}{2} =\frac{1}{3}log_{2}2^{-1}}=-\frac{1}{3}$

Szczegółowe wyjaśnienie:

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