How to get the results of this logarithmic equation? How to solve this for x: log_x(x^3+1)* log_(x+1)(x)>2 I have tried to get the same exponent by getting the second multiplier to reciprocal and tried to simplify (x^3+1).

Aydin Jarvis

Aydin Jarvis

Answered question

2022-10-28

How to get the results of this logarithmic equation?
How to solve this for x:
log x ( x 3 + 1 ) log x + 1 ( x ) > 2
I have tried to get the same exponent by getting the second multiplier to reciprocal and tried to simplify ( x 3 + 1 )

Answer & Explanation

wlanauee

wlanauee

Beginner2022-10-29Added 17 answers

We have that
log x + 1 x = log x x log x ( x + 1 ) (1) = 1 log x ( x + 1 )
and
(2) log x ( x 3 + 1 ) = log x ( x + 1 ) + log x ( x 2 x + 1 )
Using (1) and (2) reveals that
log x ( x 3 + 1 ) log x + 1 x = ( log x ( x + 1 ) + log x ( x 2 x + 1 ) ) 1 log x ( x + 1 ) (3) = 1 + log x ( x 2 x + 1 ) log x ( x + 1 )
We note that if the right-hand side of ( 3 ) is to be greater than 2, we must have log x ( x 2 x + 1 ) log x ( x + 1 ) > 1 x 2 x + 1 x + 1 > 1 x ( x 2 ) > 0 x > 2 for real-valued solutions
Thus, we have that
log x ( x 3 + 1 ) log x + 1 x > 2 for x > 2
Taniya Melton

Taniya Melton

Beginner2022-10-30Added 5 answers

Keep in mind ( log a b ) ( log c a ) = log c b
Setting a = x , b = x 3 + 1, and c = x + 1, we have
log x ( x 3 + 1 ) log x + 1 x = log x + 1 ( x 3 + 1 )
So your inequality is equivalent to
log x + 1 ( x 3 + 1 ) > 2 x 3 + 1 > ( x + 1 ) 2 x 3 + 1 > x 2 + 2 x + 1 x 3 x 2 2 x > 0 x ( x + 1 ) ( x 2 ) > 0
Then 1 < x < 0 or x > 2. One case is ruled out.

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