Odpowiedź:
[tex]\boxed {~~log_{4}5=a~~\land ~~log_{5}11=b~~\Rightarrow log_{2}\sqrt{11} =a\cdot b~~}[/tex]
Szczegółowe wyjaśnienie:
[tex]log_{4}5=a~~\land~~log_{5}11=b\\\\log_{2}\sqrt{11} =log_{2}11^{\frac{1}{2} }=\dfrac{1}{2} log_{2}11=log_{2^{2}}11=\\\\=log_{4}11=log_{4}5\cdot log_{5}11=a\cdot b[/tex]
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Odpowiedź:
[tex]\boxed {~~log_{4}5=a~~\land ~~log_{5}11=b~~\Rightarrow log_{2}\sqrt{11} =a\cdot b~~}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy ze wzorów:
Rozwiązujemy :
[tex]log_{4}5=a~~\land~~log_{5}11=b\\\\log_{2}\sqrt{11} =log_{2}11^{\frac{1}{2} }=\dfrac{1}{2} log_{2}11=log_{2^{2}}11=\\\\=log_{4}11=log_{4}5\cdot log_{5}11=a\cdot b[/tex]