How do you solve ${log}_{b} (2) = 0.24$ $log_b(2)=0.24$ ?

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Answer 1 · 2016-06-22T09:07:10

$b = 17.9594$ $b =17.9594$

${log}_{b} (2) = 0.24 \to 2 = b^{0.24}$ $log_b(2)=0.24->2=b^{0.24}$

Applying ${log}_{e}$ $log_e$ to both sides

${log}_{e} 2 = 0.24 {log}_{e} b$ $log_e 2=0.24 log_e b$

then

${log}_{e} b = \frac{{log}_{e} 2}{0.24} = 2.88811$ $log_e b = log_e 2/0.24 = 2.88811$

Finally

$b = e^{2.88811} = 17.9594$ $b = e^2.88811=17.9594$

How do you solve logb(2)=0.24log_b(2)=0.24?