# 1 + 2log 3 - log x = log(x+1)What is x?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll impose the constraints of existence of logarithms:

x>0

and

x+1>0

x>-1

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:

x*(x+1)=9*10

We'll remove the brackets and we'll move all terms to one side:

x^2 +x -90=0
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# 1 + 2log 3 - log x = log(x+1)What is x?

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll impose the constraints of existence of logarithms:

x>0

and

x+1>0

x>-1

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:

x*(x+1)=9*10

We'll remove the brackets and we'll move all terms to one side:

x^2 +x -90=0
user
The text passage contains a mathematical question and a detailed solution to that question.

TYPE: SINGLE_QUESTION_WITH_ANSWER