How do you solve #0.73^x=0.25#?

1 Answer
EZ as pi
Oct 29, 2016

#x = 4.405#

Explanation:

#0.25# is most certainly not one of the powers of #0.73#

The fact that the variable is in the index is an indication that you need to use logs.

#log0.73^x = log0.25" "larr# log both sides

#xlog0.73 = log0.25" "larr# power rule

#x = log0.25/log0.73" "larr# isolate #x#

#x = 4.405" "larr# use a calculator to get the value of #x#

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