Log

if
log⁡4x+log⁡8x=log⁡bx\log4 x+\log8 x = \log_b x
, what is b?

Result

b =  2.3

Solution:

b=411+log(4)log(8)=2.3b = 4^ { \dfrac{ 1}{1+\dfrac{ log(4)}{log(8)}}} = 2.3

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