How do you evaluate #log_9 0.4# using the change of base formula?

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Answer 1 · 2018-04-17T20:13:45

#log_9(0.4)=ln(0.4)/ln(9)~~-0.4170218835#

for:

#y=log_9(0.4)<=>9^y=0.4#

#9^y=0.4#

Taking natural logarithms of both sides:

#yln(9)=ln(0.4)#

Dividing by #ln(9)#

#y=ln(0.4)/ln(9)#

But:

#y=log_9(0.4)#

#:.#

#log_9(0.4)=ln(0.4)/ln(9)#

This is the change of base formula.

#log_9(0.4)=ln(0.4)/ln(9)~~-0.4170218835#