How do you solve #log_b(2)=0.24#? Precalculus Properties of Logarithmic Functions Functions with Base b 1 Answer Cesareo R. Jun 22, 2016 #b =17.9594# Explanation: #log_b(2)=0.24->2=b^{0.24}# Applying #log_e# to both sides #log_e 2=0.24 log_e b# then #log_e b = log_e 2/0.24 = 2.88811# Finally #b = e^2.88811=17.9594# Answer link Related questions What is the exponential form of #log_b 35=3#? What is the product rule of logarithms? What is the quotient rule of logarithms? What is the exponent rule of logarithms? What is #log_b 1#? What are some identity rules for logarithms? What is #log_b b^x#? What is the reciprocal of #log_b a#? What does a logarithmic function look like? How do I graph logarithmic functions on a TI-84? See all questions in Functions with Base b Impact of this question 1510 views around the world You can reuse this answer Creative Commons License