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hallo, please help with this:

 

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

 

Thank you very very much!!

Guest Feb 21, 2018
 #1
avatar+20009 
+1

hallo, please help with this:

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

 

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

 

laugh

heureka  Feb 21, 2018
 #2
avatar
+1

Heureka,

 

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

Guest Feb 21, 2018
 #3
avatar+88891 
+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

 

 

cool cool cool

CPhill  Feb 21, 2018

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