log(x+1)-log9 = 1-log(x)

We know that logx-logy= log(x/y)

and log10 =1

Substitute:

log(x+1)/9 = log10-log(x)

log(x+1)/9 = log(10/x)

==> (x+1)/9 = 10/x

==> x(x+1)= 90

==> x^2+x -90=0

Factorize:

(x+10)(x-9)=0

x1=-10 which is impossible because logx ca not be negative.

x2= 9

To solve

lg(x + 1) - lg9 = 1 - lg(x) .

Solution:

Using logarithm property

lga+lgb = lgab and lg(a/b) = lga-lgb, we get:

lg x+1) -lg9 = 1-lgx. Or

lg{(x+1)/9 } = lg (10/x), as lg is a notation for logarithm of base 10.

Taking antilogarithms, we get:

(x+1)/9 = 10/x. Multiplying by 9x,

x(x+1) = 90. Or

x^2+x-90 = 0. Or

x^2+10x -9x-90 =0

x(x+10) -9(x+10) = 0.

(x+10)(x-9) = 0.

x+10 = 0 or x= -10 is not practical as lg is not real for negative numbers.

x-9 =0 Or x = 9 is the real solution

We'll use the quotient property of logarithms, knowing that 1=lg10

lgx-lgy=lg(x/y)

x>0

Now, let's solve the equation:

lg(x+1)-lg9=lg[(x+1)/9]

1-lgx=lg10-lgx=lg(10/x)

lg [(x+1)/9]= lg(10/x)

We'll use the one to one property of logarithms:

(x+1)/9=10/x

We'll use the cross multiplying:

x*(x+1)=9*10

x^2 +x -90=0

We'll apply the quadratic formula:

x1=[-1+ sqrt(1+4*90)]/2=(-1+19)/2=9

x2=(-1-19)/2=-10

From the existence condition of the logarithm, x>0, so the only accepted solution is x1=9.