What is the inverse of #log (x/2)# ?

1 Answer
Mar 28, 2016

Answer:

Assuming this is base-10 logarithm, the inverse function is
#y=2*10^x#

Explanation:

Function #y=g(x)# is called inversed to function #y=f(x)# if and only if
#g(f(x))=x# and #f(g(x))=x#

Just as a refreshment on logarithms, the definition is:
#log_b(a)=c# (for #a>0# and #b>0#)
if and only if #a=b^c#.
Here #b# is called a base of a logarithm, #a# - its argument and #c# - its balue.

This particular problem uses #log()# without explicit specification of the base, in which case, traditionally, base-10 is implied. Otherwise the notation #log_2()# would be used for base-2 logarithms and #ln()# would be used for base-#e# (natural) logarithms.

When #f(x)=log(x/2)# and #g(x)=2*10^x# we have:
#g(f(x))=2*10^(log(x/2))=2*x/2=x#
#f(g(x))=log((2*10^x)/2)=log(10^x)=x#