POMÓŻCIE :)).... Question from @malinowatrusk

Janek191

z.1

= 3, bo [ p(3)]^3 = 3 p(3)

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p(3) - pierwiastek kwadratowy z 3

log 9 / log 27 = log 3^2 / log 3^3 = ( 2 log 3) / ( 3 log 3) = 2/3

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log 5 + log 16 - 3 log 2 = log 5 + log 16 - log 2^3 = log 5 + log 16 - log 8 =

= log (5*16) - log 8 = log [ (5*16)/8 ] = log 10 = 1

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4 log 3 [ 1/3] + 2 log 3 [ 9] = 4* (-1) + log 3 [ 9 ^2 ] = - 4 + log 3 [81 ] = - 4 + 4 = 0

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log 7 [ 8 ] : log 7 [ 1/8 ] = log 7 [ 8 ] : log 7 [ 8 ^(-1)] = log 7 [ 8] : ( - log 7 [ 8 ] = - 1

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[ 5^(-2) + 5^0]^(-1) = [ 1/( 5^2) + 1 ]^(-1) = [ 1/25 + 25/25]^(-1) = ( 26/25)^(-1) =

= 25/26

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64 ^(- 4/3) = [ 64 ^(1/3)]^( -4) = 4^(-4) = 1 /4^4 = 1/256

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0 votes Thanks 1

kuba303030

a) $log_{\sqrt{3}}3\sqrt{3}=3, bo (\sqrt{3})^3=3\sqrt{3}$

b) $\frac{log9}{log27}=\frac{log3^2}{log3^3}=\frac{2log3}{3log3}=\frac{2}{3}$

c) $log5+log16-3log2=log(5*16)-log2^3=log80-log8=\\=log\frac{80}{8}=log10=1$

d) $4log_3\frac{1}{3}+2log_39=4*(-1)+2*2=-4+4=0$

e) $\frac{log_78}{log_7\frac{1}{8}}=log_{\frac{1}{8}}8=-1$

f) $(5^{-2}+5^0)^{-1}=(\frac{1}{25}+1)^{-1}=(\frac{26}{25})^{-1}=\frac{25}{26}$

g) $64^{-\frac{4}{3}}=(\frac{1}{64})^{\frac{4}{3}}=\sqrt[3]{(\frac{1}{64})}^4=(\frac{1}{4})^4=\frac{1}{256}$

1 votes Thanks 2