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

1 Answer
Karthik G
Dec 24, 2015

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.

Explanation:

#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.

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