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- mathsyperson
- 2005-11-12 10:33:47
Using log a + log b = log ab, log6 2 + log6 3 = log6 6.
loga b is defined as the number that a needs to be taken to the power of to make b.
6 needs to be taken to the power of 1 to make 6, so log6 6 = 1.
- pman
- 2005-11-12 08:44:08
ahh sry my fault but how? can you explain it pls
- mathsyperson
- 2005-11-12 08:33:34
For 3, I worked it out to be 16 but typed 12 by mistake. I've edited it, so it's right now.
For 4, I said it was 1.
- pman
- 2005-11-12 08:21:22
thanx alot, but in my book it says 3)=16 and 4)=1
any ideas, and also i forogot to mention that log6 2 means the base is 6..
thx!
- mathsyperson
- 2005-11-12 06:05:32
1) lg a + lg b ≡ lg ab, so lg (1-x) + lg(1+x) = lg (1-x)(1+x) = lg (1 - x²).
lg 1 - x² = lg 0.75
1 - x² = 0.75
x² = 0.25
x = √0.25 = 0.5
2) a lg b ≡ lg b^a, so 2 lg x = lg x²
lg x² = lg 2 + lg 18 = lg 36
x⊃ 2 = 36
x = 6
3) lg a - lg b ≡ lg (a/b), so 2 lg x - lg 4 = lg (x²/4)
lg (x²/4) = lg (3x + 16)
x²/4 = 3x + 16
x² - 12x - 64 = 0
(x+4)(x-16) = 0
x = 16, because you can't have negative logarithms.
4) log6 2 + log6 3 = log6 6 = 1, because 6^1 = 6
- pman
- 2005-11-12 04:58:10
hi!
please, i need help with:
Code:
1) lg(1-x)+lg(1+x)=lg0.75 2) 2 lgx=lg2 +lg18 3) 2 lgx - lg4 = lg(3x + 16) 4) log6 2 +log6 3
thanx for the help