Oblicz:
log[5](3x) - log[5](4y)
log[6]x - 2log[6](3y)
3log[2]5+1/2log[2]a
log[3]5 * log[3]27
log[121]6 * log[36]1/11
log[(pierwiastek z 3) / 3] 4 * log[32]9
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log[5](3x/4y)
log[6](x)-log[6](3y)²=log[6](x/(3y)²)
log[2](5³)+log[2](a^½)=log[2](125·(√a))=log[2](125√a)
log[3](5)·log[3](3³)=log[3](5)·3log[3](3)=log[3](5)·3=3log[3](5)
log[11²](6)·log[6²](1/11)=(1/2)log[11](6)·(1/2)log[6](1/11)=(1/4)log[11](6)·log[6](11⁻¹) =(1/4)log[11](6)·(-1)log[6](11)=-(1/4)·1=-(1/4)
log[√3/3](4)·log[32](9) = log[√3/3](2²)·log[2⁵](3²)=2log[√3/3](2)·(1/5)·2log[2](3)=(4/5)log[√3/3](2)·log[2](3)=(4/5)log[√3/3](3)=(4/5)log[3^(-(1/2))](3)=(4/5)·(-2)log[3](3)=-8/5
mam nadzieję, że oto chodziło