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# Logs

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$$log_xa-log_{1 \over x}a^3$$

Thank you very very much!!

Feb 21, 2018

#1
+21978
+1

$$\large{ \log_x(a)-\log_{\frac1x}(a^3) }$$

$$\begin{array}{|rcll|} \hline && \mathbf{ \large{ \log_x(a)-\log_{\frac1x}(a^3) } } \\ &\large{=}& \large{ \log_x(a)-3\times \log_{\frac1x}(a) } \qquad \boxed{ \log_{\frac1x}(a) = \dfrac{\log_x(a)}{\log_x(\frac{1}{x})} } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{\log_x(\frac{1}{x})} } \qquad \boxed{ \log_x\left(\frac{1}{x}\right) = \log_x(1) - \log_x(x) } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ \log_x(1) - \log_x(x) } } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ \log_x(1) - \log_x(x^1) } } \\ && \qquad | \quad \log_x(1) =\log_x(x^0)= 0 \\ && \qquad | \quad \log_x(x^1) = 1 \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ 0 - 1 } } \\ &\large{=}& \large{ \log_x(a)+3\times \log_x(a) } \\ &\mathbf{\large{=}}& \mathbf{\large{ 4\times \log_x(a) } }\\ \hline \end{array}$$

Feb 21, 2018
#2
+1

Heureka,

thank you very much, I honestly appreciate your time!!

Guest Feb 21, 2018
#3
+99586
+1

logx a  -  log1/x a^3

Using the change-of -base  rule, we have that

log a  / log x    -    log a  /  log (1/x)   =

log a / log x  -  log a^3  / log x^-1   =

log a / log x  -  3log a / [ - log x]   =

log a / log x   +  3log a   /log x  =

4 [log a  /  log x ]  =      [ reverse the change-of-base rule ]

4 logx a

Feb 21, 2018