log (x +1) + log (x) = 1.3?

Question

Please explain your answer

Answer 1

log (x+1) + log (x) = log [(x+1)*x] = 1.3

x^2 + x = 10^1.3

Solve for x using the quadratic equation.

Answer 2

Since you didn't say, I assume all logs are base 10.

log (x +1) + log (x) = 1.3
log {(x +1)x} = 1.3
(x +1)x = 10^(1.3)
x + x² - 10^(1.3) = 0
x² + x - 10^(1.3) = 0
x = {-1 ± √[1² + 4*10^(1.3)]}/2

The negative solution is rejected since you can't take a log of a negative number.

x = {-1 + √[1 + 4*10^(1.3)]}/2 ≈ 3.9947328

Answer 3

log (x +1) + log (x) = 1.3
log x(x+1)= 1.3 (log property, i don't remember the name)
10^1.3= x^2+x
x^2+x-10^1.3

use quadractic fomula and you will get 3.995 and -4.995
you can't have a negative anwers so 3.995 is the answer

Answer 4

log (x +1) + log (x) = 1.3
=> log (x^2 + x) = 1.3
=> x^2 + x = 10^(1.3)
=> x = [1 +/- √(1 + 4.10^(1.3))] / 2
=> x = 1/2 +/- √(1/4 + 10^(1.3))
or x = 4.995, -3.995 to 3 d.p.
But we need x > 0 for log x to be defined, so we reject the second answer.
So x = 1/2 + √(1/4 + 10^(1.3)) = 4.995 (3 d.p.)

Answer 5

log(x(x+1)) = 1.3
x(x+1) = 10^1.3
x^2 + x - 10^1.3 = 0
you cant use the calculator to get
x = -4.99, 3.99
but reject -4.99 because you cant have a neg. log
so final answer...x = 3.995

Answer 6

By guessing log= 1 and X=.15. so 1(.15+1)+1(.15)=1.3

But i am just guessing and prolly just making it up