Expand the expression ln 2y/x
How does a person go about expanding that thing?
Are you working on series and sequences?
For example,
ln(1+x) = x - (x^2)/2 + (x^3)/3 -(x^4)/4... where |x|<1
or
ln(x) = (x-1) - ((x-1)^2)/2 + ((x-1)^3)/3 -.... where 0<x≤2
etc.
I don't think we are working on series and sequences becuase that stuff doesnt look familiar... we are just doing logorithms and ln stuff... is there any other way you can expand it?
thanks!
Shikra
In that case, it would require the laws of logarithms, such as:
log(a)+log(b) = log(ab)
log(a)-log(b) = log(a/b)
b*log(a) = log(ab)
log(a)/b = log(a^(1/b))
...
etc.
Expand 2y/x into 2*y÷x
and use the above laws to expand the logarithm part.
To expand the expression ln(2y/x), you can use the properties of logarithms.
The natural logarithm ln(x) is the inverse of the exponential function e^x. So, ln(2y/x) represents the power to which e must be raised to obtain (2y/x), or in other words, it represents the exponent.
To expand the expression, we can use the logarithmic rule:
ln(ab) = ln(a) + ln(b)
Applying this rule to ln(2y/x), we can rewrite it as:
ln(2) + ln(y) - ln(x)
So, the expanded form of ln(2y/x) is ln(2) + ln(y) - ln(x).