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Stiind ca log in baza 3 din 2=a
Demonstrati ca log in baza 16 din 24=(1+3a)/4a


Răspuns :

[tex]a = log_{3}(2) [/tex]

[tex] log_{16}(24) = \frac{1 + 3a}{4a} [/tex]

[tex] log_{16}(24) = \frac{ log_{3}(3) + 3 \times log_{3}(2) }{4 \times log_{3}(2) } [/tex]

[tex] log_{ {2}^{4} }(3 \times {2}^{3} ) = \frac{ log_{3}(3) + log_{3}( {2}^{3} ) }{ log_{3}( {2}^{4} ) } [/tex]

[tex] log_{ {2}^{4} }(3 \times {2}^{3} ) = \frac{ log_{3}(3 \times {2}^{3} ) }{ log_{3}( {2}^{4} ) } [/tex]

[tex] log_{ {2}^{4} }(3 \times {2}^{3} ) = log_{ {2}^{4} }(3 \times {2}^{3} ) [/tex]

[tex] log_{16}(24) = log_{16}(24) \: (A)[/tex]

Formule folosite :

[tex]1 = log_{x}(x) [/tex]

[tex]log_{x}(y)+log_{x}(z)=log_{x}(y\:\times\:z)[/tex]

[tex]n \times log_{x}(y) = log_{x}( {y}^{n} ) [/tex]

[tex] log_{x}(y) = \frac{ log_{z}(y) }{ log_{z}(x) } [/tex]