How do you solve #log_b(2) = .105#?

Question

Chloe Alexander · Accepted Answer

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.

#log_b(2) = 0.105# The rule #log_b(a) = k => a=b^k# Using this rule we get # 2 = b^0.105# We can solve this by taking #0.105# root of #2# #2^(1/0.105) = b# #b =736.12630909184909714688332138981# using calculator. Alternate Method #log_b(2) = 0.105# Using change of base rule which says #log_b(a) = log(a)/log(b)# We get #log(2)/log(b) = 0.105# Cross multiplying we get #log(2) = 0.105*log(b)# #log(2)/0.105 = log(b)# #2.8669523396569637639403704259476 = log(b)# #b = 10^2.8669523396569637639403704259476 #b=736.12630909184909714688332138989# We can round it as per requirement or instructions.

Sadie Ackerman · Answer

undefined To solve the equation $ \log_b(2) = 0.105 $, you can use the property of logarithms that states: $ \log_b(x) = y $ is equivalent to $ x = b^y $. Therefore, to solve for $ b $, you raise the base $ b $ to the power of $ 0.105 $. This can be expressed as:

$$ b = 2^{0.105} $$

By evaluating $ 2^{0.105} $, you can find the value of $ b $.