How do you solve #log_b2 = .3562#?

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Answer 1 · 2016-08-30T11:22:24

#b =7.00027 #

#log_b 2=log_e2/(log_e b) = .3562#

so

#log_eb=(log_e2)/.3562 = log_e 2^(1/.3562)#

finally

#b = e^( log_e 2^(1/.3562))=2^(1/.3562) = 7.00027#

Answer 2 · 2016-08-30T11:31:58

#b~=7#

#log_b 2 = 0.3562#

#2= b^0.3562#

#b= 2^(1/0.3562) ~= 2^2.807412#

#b~= 7.00028 ~= 7 #

To check result:

#log_7 2 = ln2/ln7 ~= 0.69315/1.94591 #

# ~= 0.3562#