How do I solve this: lg (1-x) + lg (1+x) = lg0,75?

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canoneuser

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Hello, I am wondering about this one: How do I solve this: lg (1-x) + lg (1+x) = lg0,75

What I did was this:
lg1 - lgx + lg1 + lgx = lg0,75
lg 1/x + lgx = lg0,75

Is this right :P
Thank you!
 
Sorry, but you are way off.

First, log(1-x) ? log(1) - log(x)

Remember this rule:
log(a*b) = log(a) + log(b)

You have
log(1-x) + log(1+x) = log(c) where c is a constant [equal to 0,75 in this particular case]

Using the rule I mentioned above you can write the left hand side as:
log[(1-x)(1+x)] = log(c)
expand what's in the square brackets to get:
log(1 - x^2) = log(c)

Now you have log's on both sides and if they are to be equal then their arguments have to be equal. This means
(1 - x^2) = c

Rewrite this as:
x^2 - 1 + c = 0
substitute 0,75 for c to get
x^2 - 0,25 = 0

You should recognize the form on the left as something you can easily factor.
(x - 0,5)(x + 0,5) = 0

so x = ± 0,5
 
Sorry, but you are way off.

First, log(1-x) ? log(1) - log(x)

Remember this rule:
log(a*b) = log(a) + log(b)

You have
log(1-x) + log(1+x) = log(c) where c is a constant [equal to 0,75 in this particular case]

Using the rule I mentioned above you can write the left hand side as:
log[(1-x)(1+x)] = log(c)
expand what's in the square brackets to get:
log(1 - x^2) = log(c)

Now you have log's on both sides and if they are to be equal then their arguments have to be equal. This means
(1 - x^2) = c

Rewrite this as:
x^2 - 1 + c = 0
substitute 0,75 for c to get
x^2 - 0,25 = 0

You should recognize the form on the left as something you can easily factor.
(x - 0,5)(x + 0,5) = 0

so x = ± 0,5
 
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