# Determine the solutions of the equation lg(x+1) - lg9 = 1 - lg(x)

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### 3 Answers

Let's use some logarithmic rules before:

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)

From one of the properties of logarithmic function, the one which says that this function is an injection:

(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 is x1=9

To solve lg(x+1)-lg9 = 1-lgx.

We move the unknowns and isolate them:

lg(x+1)+lgx =lo9+1.

lg(x+1)x = log9+log10, asĀ lga+lgb = lgab and as10^1 =1, lg10 = 1,

by definition.

lg(x+1)x = lg9*10

x(x+1) = 9*(10).

x = 9 , obviously satisfies the equation.

x = 9.

We use the relation for logarithms here: log a + log b = log ab and log a - log b = log(a/b).

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

=> lg (x+1) + lg x - lg 9 = 1

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

taking antilog of both the sides

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

=> 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 and x =-10 .

Now as lg of a negative value is not defined x =-10 is not valid .

**So x =9**