Natural Logarithmic Function

In summary: I think he meant to solve for x as follows:ln(2x+1)=2-ln(x) so, adding ln(x) to both sides, ln(2x+1)+ln(x)=ln(x(2x+1))=2. Now take the exponential of both sides: x(2x+1)=2x2+x=e2 and you can solve that quadratic equation for x.
  • #1
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Homework Statement



Solve ln(2x+1)=2-ln(x) for x.


Homework Equations





The Attempt at a Solution



e^(ln(2x+1)) = e^((2-ln(x))

2x + 1 = e^(2/x)

2x^2 + x = e^2

2x^2+x-e^2 = 0

At this point, I know you're supposed to use the quadratic equation. But, my problem is with how to treat e^2. Should I multiply it out in the discriminant where c = e^2? or just e?

Thanks for any help.
 
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  • #2
e^2 is just a number... you can call it c if you like.
 
  • #3
You can either just calculate it out (it's only a number!) or include it in the quadratic formula and quote your answers in terms of e.
It depends wether this is for home work or real work!
 
  • #4
There is an error in the second part

e^(ln(2x+1)) = e^((2-ln(x))

2x + 1 = e^(2/x)

These are NOT equivalent. Remember that you can only turn logarithmic expressions
into quotients of logarithms in a case like this: ln z - ln x = ln (z/x)

Try again using that info.
 
  • #5
BrendanH said:
There is an error in the second part

e^(ln(2x+1)) = e^((2-ln(x))

2x + 1 = e^(2/x)

These are NOT equivalent. Remember that you can only turn logarithmic expressions
into quotients of logarithms in a case like this: ln z - ln x = ln (z/x)

Try again using that info.
I suspect that was a typo because in the next line he has
2x^2+ x= e^2 which is correct.

Added Later: In fact, I would have been inclined to solve the problem as follows:
ln(2x+1)=2-ln(x) so, adding ln(x) to both sides, ln(2x+1)+ ln(x)= ln(x(2x+1))= 2.
Now take the exponetial of both sides: x(2x+1)= 2x2+ x= e2 and you can solve that quadratic equation for x.
 
Last edited by a moderator:
  • #6
BrendanH, Rossinole meant (e^2)/x, not e^(2/x). The next line in the derivation,
2x^2 + x = e^2
is correct.

edit added:
Halls beat me to it.
 

What is a natural logarithmic function?

A natural logarithmic function is a mathematical function that is used to describe the relationship between two quantities, where the rate of change of one quantity is proportional to the value of the other quantity. It is also known as the "logarithm with base e" or the "ln function".

What is the base of a natural logarithmic function?

The base of a natural logarithmic function is the mathematical constant e, which is approximately equal to 2.71828. This means that the natural logarithm of a number is the power to which e must be raised to equal that number.

What is the difference between a natural logarithmic function and a common logarithmic function?

A natural logarithmic function uses a base of e, while a common logarithmic function uses a base of 10. This means that the natural logarithm of a number will be a different value than the common logarithm of the same number.

What are the properties of natural logarithmic functions?

Some of the properties of natural logarithmic functions include being the inverse function of the exponential function, having a domain of positive real numbers, and having a graph that approaches but never touches the x-axis as x approaches negative infinity.

How are natural logarithmic functions used in science?

Natural logarithmic functions are used in many areas of science, including physics, chemistry, biology, and economics. They can be used to model natural phenomena, such as population growth or radioactive decay, and to solve complex equations that involve exponential growth or decay.

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