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:

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

$~= 0.3562$

How do you solve log_b2 = .3562log_b2 = .3562?