3 szkoły ponadpodstawowej

III liceum

III technikum

MATeMAtyka 3. Zakres podstawowy

a) log6​4+log6​9=log6​(4⋅9)=log6​36=log6​62=2   <hr> b)

a) <ul> <li> 2+log3​5,   a=3  </li> </ul> 2+log3​5==2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMinYMin slice'><path d='M0 6l6-6h17c12.688 0 19.313.3 20 1 4 4 7.313 8.3 10 13
 35.313 51.3 80.813 93.8 136.5 127.5 55.688 33.7 117.188 55.8 184.5 66.5.688
 0 2 .3 4 1 18.688 2.7 76 4.3 172 5h399450v120H429l-6-1c-124.688-8-235-61.7
-331-161C60.687 138.7 32.312 99.3 7 54L0 41V6z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMidYMin slice'><path d='M199572 214
c100.7 8.3 195.3 44 280 108 55.3 42 101.7 93 139 153l9 14c2.7-4 5.7-8.7 9-14
 53.3-86.7 123.7-153 211-199 66.7-36 137.3-56.3 212-62h199568v120H200432c-178.3
 11.7-311.7 78.3-403 201-6 8-9.7 12-11 12-.7.7-6.7 1-18 1s-17.3-.3-18-1c-1.3 0
-5-4-11-12-44.7-59.3-101.3-106.3-170-141s-145.3-54.3-229-60H0V214z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMaxYMin slice'><path d='M399994 0l6 6v35l-6 11c-56 104-135.3 181.3-238 232-57.3
 28.7-117 45-179 50H-300V214h399897c43.3-7 81-15 113-26 100.7-33 179.7-91 237
-174 2.7-5 6-9 10-13 .7-1 7.3-1 20-1h17z'/></svg>log3​9​​+log3​5=log3​(9⋅5)=log3​45   <hr> b)

a) log12​2+log12​8+log12​9=log12​(2⋅8⋅9)=log12​144=log12​122=2   <hr> b)

Z treści zadania wiemy, że log5≈0,7   <hr> a)

Z treści zadania wiemy, że log2​7≈2,8   <hr> a)

Z treści zadania wiemy, że <ul> <li> log5​2≈0,43 </li> <li> log5​7≈1,21   </li> </ul> <hr> a)

a) p=log2​5+log2​6=log2​(5⋅6)=log2​30   <hr> b)

a) log4​2+log4​8=log4​(2⋅8)=log4​16=log4​42=2   <hr> b)

a) log4​12−log4​3=log4​ 312​=log4​4=1   <hr> b)

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