You are not logged in.
Pages: 1
hi!
please, i need help with:
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
Offline
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
Why did the vector cross the road?
It wanted to be normal.
Offline
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!
Offline
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.
Why did the vector cross the road?
It wanted to be normal.
Offline
ahh sry my fault but how? can you explain it pls
Offline
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.
Why did the vector cross the road?
It wanted to be normal.
Offline
Pages: 1