log(1-x)-log(1+x)=88?

Question

solve show work

Answer 1

log(1-x)-log(1+x)=88
log((1-x)/(1+x)) = 88
(1-x)/(1+x) = 10^88
1-x = 10^88 * (1+x)
x(10^88 + 1) = (1 - 10^88)
x = (1 - 10^88) / (1 + 10^88)

That's very close to -1, -0.9999999999999999999999999999999[etc.]

Answer 2

Since the base is not shown, we will assume that these are base 10 logarithms or common logarithms.

log(1 - x) - log(1 + x) = 88

log[ (1 - x) / (1 + x ) ] = 88

10^88 = (1 - x) / ( 1 + x)

10^88 + (10^88)x = 1 - x

(10^88)x + x = 1 - 10^88

x(1 + 10^88) = 1 - 10^88

x = (1 - 10^88) / (1 + 10^88)

x is a negative number that is very close -1, but larger than -1.

Answer 3

log(1-x)-log(1+x)=88

Using log(a)-log(b) = log(a/b) formula we get,

log((1-x)/(1+x)) = 88

Generally log(x) means log(x) to the base 10. So if log(x) = k, then it means x = 10^k. Applying this principle we get,

(1-x)/(1+x) = 10^88 (is a very big number). Lets assume this to be 'k'.

Therefore (1-x)/(1+x) = k = 10^88.

=> 1-x = k(1+x) = k + kx
=> kx + x = 1-k
=> x*(k+1) = 1-k
=> x = (1-k)/(1+k)

i.e x = (1 - 10^88)/(1 + 10^88)

Instead of 10 if the logarithmic base is assumed to some constant 'b', then the general solution to the problem will be:

x = (1 - b^88)/(1 + b^88)

Answer 4

Since log a + log b = log ab, you can write:

log (1-x/ 1 +x) = 88

Take 10 to the power of each side.

(1-x/1+x) = 10^88

You can solve that.

Answer 5

log (1-x)- log (1+x)=88
(1-x)/(1+x)=10^88
1-x=(10^88)+(10^88)x
1-10^88=(10^88)x+x
1-10^88=x(10^88 +1)
x=(1-10^88)/(1+10^88)

Answer 6

log(1-x) - log(1+x) = 88
log[(1-x)/(1+x)] = 88
(1-x)/(1+x) = e^88
1-x = e^88 + e^88*x ....(here * means product)
(e^88 - 1)x = (1 - e^88)
x = (1 - e^88)/(e^88 - 1)

Answer 7

Remember the properties of Logarithms

log(xy) = log(x) + log(y) (multiplication rule)
log(x/y) = log(x)-log(y) (division rule)

Use one of these rules.

Also when you get the single logarithm on the left side
raise the left side to the natural base (e) to undo the logarithm
and raise the right side to base (e). Then solve for x.