How do you evaluate #f(x)=2lnx# for #x=0.75#?

1 Answer
LizZ
Sep 3, 2017

With a calculator. And #f(0.75)~~-.5754#

Explanation:

You can do all sorts of things to this expression, but nothing will get you closer to an answer without using a calculator.
For example, on the right side, you have a rule of logarithms that says you can move that #2# in front of the natural log sign to the place of an exponent of the #x#:
#2ln 0.75= ln 0.75^2#.

You can rewrite #y= ln 0.75^2# as #e^y=0.75^2#, but this still gets you no closer to getting an answer.

So, plug it in your calculator. And if you don't understand what #e# is, or what logarithms are, check out videos about logarithms on Khan Academy and Eddie Woo's [video] on #e#.
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