# Can I multiply e to ln to cancel ln?

For example, I want to cancel 2ln on both sides of the equation': 2ln(√x+1)=2ln(√y+1) by multiplying all to 1/2 e. Is that correct?

Relevance

No, if you want to get rid of the ln

You raise e^() of both sides

e^(2(ln( sqrt(x) +1)  = e^(2*ln (sqrt(y) + 1))

e^(ln  ( (sqrt(x) + 1)^2 ) =   e^(ln ( sqrt(Y) +1)^2 )

(sqrt(x) + 1)^2  = (sqrt(y) + 1) ^2

or you could multiply both side by 1/2 first

then raise to e^() of both sides

e^(ln( sqrt(x) + 1)  ) = e^(ln(sqrt(y) + 1)

sqrt(x) + 1 = sqrt(y) +1

sqrt(x) =sqrt(y)   x> 0 , y> 0

square both sides

x = y     for x> 0  and y >0

• You cannot. There is no meaning in writing elnx = x. Because "ln" is a function. Not a number.

• (√x+1) = e^(ln(√x+1))

ln(...) is a function that produces a unique value.

So, it's like you have f(√x+1) = f(√y+1)

where f(...) is a 1 to 1 function

in which case, you can conclude that

(√x+1) = (√x+1)

• 2ln(√x+1)=2ln(√y+1)

First remove the coefficient '2' to become the index number.

Hence

2ln(√x+1)=2ln(√y+1)

Then Antilog.

(√x+1)^2=(√y+1)^2

• If you double the logarithm of a number, it becomes the logarithm of the SQUARE of the number.  That is, if, say k = ln(z), then 2k = ln(z^2).

2ln(√x+1)=2ln(√y+1)

to re-write it as

ln(√x+1)^2 = ln(√y+1)^2

and if the logarithms are equal, then the numbers must be equal, that is

(√x+1)^2 = (√y+1)^2

Is that what you are looking for?

(Of course, this obviously means that x = y).

• requires a reverse logarithm, which is an exponent.  you need to use e^x to get rid of the ln x.  e^(lnx) actually.  need both sides to have the same change to maintain the equality, of course, and I am sure you already knew that.

So, you might change a problem of lnx on one side into an e^x (or some other variable) on the other.

• NO.  ln is an operation -- like multiplication. division, addition, and subtraction are math operations....

. . .e  on the other hand is a number .. not an operation

just like pi is a number (approx. 3.14159)  e is also an irrational number approx 2.71828 = = = ln is NOT a number but an operation preformed on on a number ===  ln (4)  =  1.38629 (approx)  = = =you are not multiplying by ln

• Yes, you can get rid of the 2's by dividing by 2 (multiplying by half). Then you get ln operating on the whole of both sides. Mathematically you remove logs by "taking logs" but here just by reason, if it operates on the whole of each side and they are equal then they must have been the same before the operator was applied, so you can just remove it. In fact by inspection you can just look at the whole of

2ln(√x + 1) = 2ln(√y + 1)

and see by symmetry that to be equal, then x = y, with some cavitats about ±