1 + 2log 3 - log x = log(x+1)What is x?
First, we'll impose the constraints of existence of logarithms:
The common interval of admissible values for x is (0 , +inf.).
We'll re-write the term 2log 3 using the power rule of logarithms:
2log 3 = log 3^2
2log 3 = log 9
We'll re-write the equation, moving all terms that contain the variable x to one side:
1 + log 9 = log(x+1) + log x
Now, we'll solve the equation, applying th product rule of the logarithms:
log9 + log 10 = lg(x+1) + lgx
log 9*10 = log[x*(x+1)]
Because the bases of logarithms are matching, we'll use the one to one property:
We'll remove the brackets and we'll move all terms to one side:
x^2 +x -90=0
We'll apply the quadratic formula:
x1=[-1+ sqrt(1+4*90)]/2=(-1+19)/2=9 > 0
x2=(-1-19)/2=-10 < 0
We'll reject the second solution and the equation will have just the solution x = 9.