Determine the solutions of the equation lg(x+1) - lg9 = 1 - lg(x)
Let's use some logarithmic rules before:
Now, let's solve the equation:
From one of the properties of logarithmic function, the one which says that this function is an injection:
We'll use the cross multiplying:
x^2 +x -90=0
x1=[-1+ sq root(1+4*90)]/2=(-1+19)/2=9
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)x = log9+log10, as lga+lgb = lgab and as10^1 =1, lg10 = 1,
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