How do you solve #log_b(2)=0.24#?

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Answer 1 · 2016-06-22T09:07:10

#b =17.9594#

#log_b(2)=0.24->2=b^{0.24}#

Applying #log_e# to both sides

#log_e 2=0.24 log_e b#

then

#log_e b = log_e 2/0.24 = 2.88811#

Finally

#b = e^2.88811=17.9594#