How do you solve #log_b(2) = .105#?
The trick here would be to convert the log function to exponent form and then solve. Please check the explanation for two approaches to solve the same.
Using this rule we get
We can solve this by taking
Using change of base rule which says
Cross multiplying we get
#b = 10^2.8669523396569637639403704259476
We can round it as per requirement or instructions.