How do you subtract logs with different bases?
2007-07-09 18:52:45 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Logs with different bases are divided by converting them into common base....
3log(base 3)81
=3log(base 3)3^4
=12log(base3)3
=12
log(base 4)8 + log(base 4)128
=Log(base4)(8*128)
=log(base4)1024
=log(base4)4^5
=5log(base4)4
=5
3log(base 3)81 - log(base 4)8 - log(base 4)128
=3log(base 3)81 - {log(base 4)8 + log(base 4)128}
=12-5
=7
TW K
2007-07-09 19:03:46 · answer #1 · answered by TW K 7 · 1⤊ 0⤋
3log(base 3)81 - log(base 4)8 - log(base 4)128
= 3log(base 3)3^4 - log(base 4)2^3 - log(base 4)2^7
= 12log(base 3)3 - 3log(base 4)2 - 7log(base 4)2
= 12 - 10log(base 4)2
= 12 - 10log(base 4)4^(0.5)
= 12 - 10(0.5)
= 12 - 5
= 7
Alt, u can use tis formula: to change base a to base b,
log(base a)N = [log(base b)N] / [log(base b)a]
2007-07-09 20:39:33 · answer #2 · answered by Anonymous · 0⤊ 0⤋
3log(base 3)81 - log(base 4)8 - log(base 4)128
= 3log(base 3)3^4 - log(base 4)4^(1.5) - log(base 4)4^(3.5)
= 3*4 - 1.5 - 3.5 = 12 - 1.5 - 3.5 = 7
2007-07-09 19:00:50 · answer #3 · answered by Northstar 7 · 0⤊ 0⤋
3log(base 3)81 - log(base 4)8 - log(base 4)128
=3log(base3)3^4 - log(base 4)(8*128)
since loga+logb=log(a*b) if base are same
=3*4log(base3)3-log(base4)(4^5)
since loga^b=b loga
=12-5log(base4)4
since log (basea) a = 1
=12-5
=7
2007-07-09 18:59:07 · answer #4 · answered by Jain 4 · 0⤊ 0⤋
= 3 x 4 - [ log4 (8) + log4 (128) ]
= 12 - log4 (1024)
= 12 - 5
= 7
2007-07-13 18:37:10 · answer #5 · answered by Como 7 · 0⤊ 0⤋
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2016-11-08 21:25:51 · answer #6 · answered by deller 4 · 0⤊ 0⤋
12-1.5-3.5=7
2007-07-09 18:55:49 · answer #7 · answered by Nishant P 4 · 0⤊ 1⤋