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**pman****Member**- Registered: 2005-11-03
- Posts: 10

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

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**pman****Member**- Registered: 2005-11-03
- Posts: 10

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!

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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**pman****Member**- Registered: 2005-11-03
- Posts: 10

ahh sry my fault but how? can you explain it pls

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

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