How do you evaluate #log_(1/2) 5# using the change of base formula?

1 Answer
Gió
Sep 16, 2017

I got: #-2.32193#

Explanation:

Consider a log such as:

#log_ba#

we can change into the new base #e# as:

#log_ba=color(red)((log_ea)/(log_eb))#

as you can see the new base gives a fraction between two new logs BUT we can easily use our calculator to evaluate #log_e# that is called the Natural Logarithm and is indicated as #ln#. So we get:

#log_ba=(log_ea)/(log_eb)=ln(a)/ln(b)=#

in our case:

#ln(5)/ln(1/2)=(1.60943)/(-0.69315)=-2.32193#

[I forgot, you can test your result by applying the definition of log:
we got that:
#log_(1/2)(5)=-2.32193#

so:
#(1/2)^(-2.32193)=5#
that can be evaluated with the calculator]

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