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

we know that log x -log y = logx/y

we also know that log10 = 1

Let us substitute"

==> log(x+1) - log 9 = log 10 -log x

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

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

==> x(x+1) = 90

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

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

==> x= 9

(we will not consider x= -10 because the function is not defined for negative values)

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

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

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

x+1/9 = 10/x

x^2 + x = 90

x^2 + x - 90 = 0

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

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

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

x = 9 or x = -10

if we take x = -10

then logx = log(-10) which is impossible

so x = 9

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

Solution:

We use ln for logarithm for base e and lg for logarithms of base 10.

Logarithm rules are lga = lg b = lg(a/b) and lga + lgb = lg(ab).

So, LHS = lg(x+1)- lg9 = lg [(x+1)/9] = RHS = 1- lg (x) Or

lg[(x+1)/9]+lgx = 1. Or

lg{(x+1)x/9 = 1. Taking anti logarithms (with respect to the base 10, w e get:

(x+1)x/9 = 10^1 . Or

(x+1)x = 10*9. Obviously and seemingly x= 9 by comparison.

x^2+x-90 = 0.This is quadratic which could be factorised and solved for x:

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

x-9 = 0 gives x =9 which is a practical solution.

x +10 = 0 gives x = -10, which is not practical as logarithm of negative does not exist.

Let's recall the logarithmic properties before:

1 = lg10

The quotient law: 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)

From one to one property of logarithmic functions, we'll get:

(x + 1)/9 = 10/x

We'll use the cross multiplying:

x*(x + 1) = 9*10

x^2 + x - 90 = 0

x1 = [-1+ sq root(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 of the equation is x1 = 9.