How do you evaluate ${log}_{20} 0.125$ $log_20 0.125$ using the change of base formula?

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Answer 1 · 2018-04-15T16:14:22

${log}_{20} 0.125 = - 0.6941$ $log_20 0.125=-0.6941$

According to change of base formula

${log}_{a} b = \frac{{log}_{x} b}{{log}_{x} a} = \frac{log b}{log a}$ $log_ab=log_xb/log_xa=logb/loga$

and when we use base $10$ $10$ , we do not write $10$ $10$ in subscript.

Hence ${log}_{20} 0.125$ $log_20 0.125$

= $\frac{log 0.125}{log 20}$ $log0.125/log20$

= $\frac{log (\frac{1}{8})}{log (2 \cdot 10)}$ $log(1/8)/log(2*10)$

= $\frac{log 1 - log 8}{log 2 + log 10}$ $(log1-log8)/(log2+log10)$

= $\frac{- {log 2}^{3}}{log 2 + 1}$ $(-log2^3)/(log2+1)$

= $- \frac{3 log 2}{1 + log 2}$ $-(3log2)/(1+log2)$

= $- \frac{3 \cdot 0.3010}{1.3010}$ $-(3*0.3010)/1.3010$

= $- 0.6941$ $-0.6941$

How do you evaluate log200.125log_20 0.125 using the change of base formula?