Is the ln(x+1) equal to ln(x) + ln(1)?

Question

I have to finish my summer homework and i am down to the last few and i just don't know how to finish them. And the reason i am doing it last minute is not because i am procrastinating. I have had a 9-6 job, at the Goddard School which is like a daycare/preschool, all summer to help pay for my car.

But the original math question is solve the equation:

ln x = ln (x+1) - 1 and i don't evem know where to start.

Answer 1

For all cases, no. But if you do want to find the solution as to when it is true, this is how you would go about it:

ln x = ln (x+1) - 1
e^(ln x) = e^(ln (x+1) - 1)
e^(ln x) = e^(ln (x+1))/e^1
x = (x+1)/e
ex = x+1
ex-x = 1
x(e-1) = 1
x = 1/(e-1)

Answer 2

I n x = I n (x+1) -1 = I n x

Answer 3

problem 1
As per logarithmic rules
ln(ab) = ln a + ln b --- product rule
ln(a/b) = ln - ln b ----- quotient rule
So ln(x+1) is not equal to ln(x) + ln(1)
because ln(x) + ln(1) = ln(x)(1) = ln x

Problem 2
ln x = ln(x +1) -1
logarithm of any number to the base of same number =1
so 1 can be written as ln(e) to the base e
ln(x) = ln(x + 1) - ln(e)
ln(x) = ln[(x +1)/(e)] ---- applying quotient rule
taking out logarithms
x = (x + 1)/e
cross multiplying
ex = x + 1
ex - x = 1
x(e - 1) = 1
x = 1/(e - 1)

Answer 4

Generally not.
But if it is an equation, then you have a specific value of x that satisfies ln x = ln (x+1) - 1 (at x = .58197671).
-----------
Ideas to solve the equation:
Collect variable terms in one side,
ln(x+1) - ln(x) = ln(1+1/x) = lne
1+1/x = e
x = 1/(e-1) = .58197671

Answer 5

no,it's not. ln(x+1) is not equal to ln(x) = ln(1)


second problem:
ln (x) = ln (x+1) - 1

subtract ln(x + 1) for both sides
ln(x) - ln(x+1) = -1

quotient rule: ln(a) - ln(b) = ln(a/b)
ln [ x/(x+1)] = -1

raise the equations with the power or e
e^(ln [ x/(x+1)] = e^-1

memorise this rule;
e^(ln(x)) = x

x/(x+1) = 1/e

ex = x+1

ex - x = 1

x(e - 1) = 1

x = 1/(e-1)

Answer 6

no it is not. However, the ln(x*1) is equal to ln(x) +ln (1).

I would suggest doing:
lnx +1 = ln(x+1)
lnx + ln(e) = ln (x+1)
ln (ex) = ln(x+1)

and playing with it from there

Answer 7

No, it is not.

ln(x) = ln(x + 1) - 1
ln(x+1) - ln(x) = 1
ln[(x+1)/x] = 1 (ln(a) - ln(b) = ln(a/b), that you can do)
(x+1)/x = exp(1) = e
x + 1 = ex (e times x)
1 = (e -1)x
x = 1/(e - 1)

Answer 8

ln x = ln (x+1) -1

ln (x+1) - ln x = 1

ln { (x+1)/ x} = 1

(x+1)/ x = ln inv 1

1+ 1/x = 2.72
1/x = 1.72
x = 0.58

Answer 9

ln(x+1) is not equal to lnx+ln1.

lnx=ln(x+1)-1
ln[(x+1)/x]=1
(x+1)/x=e
x+1=ex

(e-1)x=1
x=1/(e-1). ANS.

Answer 10

ln x = ln(x+1) - 1
e^(ln x) = e^(ln (x+1) - 1)
x = e^(ln(x+1)*e^(-1)
x = (x+1)/e
ex = x + 1
ex - x = 1
x (e -1) = 1
x = 1/(e-1)