How do you solve ${log}_{b} 2 = .3562$ $log_b2 = .3562$ ?

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Answer 1 · 2016-08-30T11:22:24

$b = 7.00027$ $b =7.00027$

${log}_{b} 2 = \frac{{log}_{e} 2}{{log}_{e} b} = .3562$ $log_b 2=log_e2/(log_e b) = .3562$

so

${log}_{e} b = \frac{{log}_{e} 2}{.3562} = {log}_{e} 2^{\frac{1}{.3562}}$ $log_eb=(log_e2)/.3562 = log_e 2^(1/.3562)$

finally

$b = e^{{log}_{e} 2^{\frac{1}{.3562}}} = 2^{\frac{1}{.3562}} = 7.00027$ $b = e^( log_e 2^(1/.3562))=2^(1/.3562) = 7.00027$

Answer 2 · 2016-08-30T11:31:58

$b ≅ 7$ $b~=7$

${log}_{b} 2 = 0.3562$ $log_b 2 = 0.3562$

$2 = b^{0.3562}$ $2= b^0.3562$

$b = 2^{\frac{1}{0.3562}} ≅ 2^{2.807412}$ $b= 2^(1/0.3562) ~= 2^2.807412$

$b ≅ 7.00028 ≅ 7$ $b~= 7.00028 ~= 7$

To check result:

${log}_{7} 2 = \frac{ln 2}{ln 7} ≅ \frac{0.69315}{1.94591}$ $log_7 2 = ln2/ln7 ~= 0.69315/1.94591$

$≅ 0.3562$ $~= 0.3562$

How do you solve logb2=.3562log_b2 = .3562?