Răspuns:
Explicație pas cu pas:
folosim formula de schimbare a bazei:
[tex]log_a x = \frac{\Big log_b x}{\Big log_b a}[/tex]
asadar:
a)
[tex]\Big log_{25}\ 8 = \frac{\Big log_5 8}{\Big log_5 25} = \frac{\Big log_5 \sqrt{64} }{\Big log_5 5^2} = \frac{\Big log_5 \sqrt{4^3} }{2*\Big log_5 5} = \frac{\Big log_5 (4^3)^\frac{1}{2} }{2*1} = \frac{\Big log_54^\frac{3}{2} }{2} = \frac{\frac{3}{2}*\Big log_54 }{2} = \frac{3 }{4}*\Big log_54[/tex]
b) folosim formula :
[tex]log_a x*y = log_a x + log_a y[/tex]
asadar:
[tex]log_3 12 = log_3 3*2^2 = log_3 3 + log_32^2 = 1 + 2*log_32[/tex]
deci
[tex]1 + 2*log_32 = log_3 12 \\\\2*log_32 = log_3 12 - 1 \\\\log_32 = \frac{log_3 12 - 1 }{2} \\\\[/tex]
deci avem:
[tex]log_2 9 = log_2 3^2 = 2*log_23 = 2* \frac{\Big log_3 3}{\Big log_3 2} = 2* \frac{\Big 1 }{\frac{\Big {log_3 12 \ - 1} }{\Big 2}} = \frac{\Big 4}{\Big{log_3 12 \ - 1 }} \\\\[/tex]
