why does ln(1/x)= -1/x ?
is that the derivative? if it is,
ln(1/x) can be written as:
ln (x^-1) = -ln(x)
since the derivative of ln (x) is 1/x, then
derivative of -ln(x) = ln(1/x)= -1/x
hope this helps.
ohhh~ Thank You!
ln(1/x)= ln(1)-ln(x)
ln(1))=0
ln(1)-ln(x)=0-ln(x)= -ln(x)
ln(1/x)=-ln(x)
To understand why ln(1/x) equals -1/x, let's break it down step by step.
First, let's recall the definition of the natural logarithm. The natural logarithm of a number "x" is denoted as ln(x) and represents the power to which the mathematical constant "e" (approximately 2.71828) must be raised to obtain that number. In mathematical terms, ln(x) = y can be expressed as e^y = x.
Now, let's apply this definition to ln(1/x). We need to determine the power to which "e" should be raised to obtain 1/x. Mathematically, we have e^y = 1/x.
To solve for "y" in this equation, we can manipulate it further. We can multiply both sides of the equation by x to isolate the fraction:
e^y * x = 1
Next, we can rewrite 1 as e^0 since anything raised to the power of 0 is equal to 1:
e^y * x = e^0
Since the bases of the exponential terms are the same (e), we can equate the exponents:
y * x = 0
Now we want to isolate "y" on one side, so we divide both sides of the equation by "x":
y = 0/x
Finally, we simplify 0/x to obtain the power y:
y = 0
Therefore, ln(1/x) is equal to 0.
To summarize: ln(1/x) = 0, not -1/x. If you encountered a source or calculation stating that ln(1/x) equals -1/x, it is likely incorrect.