hallo, please help with this:
\(log_xa-log_{1 \over x}a^3\)
Thank you very very much!!
+26396 hallo, please help with this:
\(\large{ \log_x(a)-\log_{\frac1x}(a^3) }\)
\(\begin{array}{|rcll|} \hline && \mathbf{ \large{ \log_x(a)-\log_{\frac1x}(a^3) } } \\ &\large{=}& \large{ \log_x(a)-3\times \log_{\frac1x}(a) } \qquad \boxed{ \log_{\frac1x}(a) = \dfrac{\log_x(a)}{\log_x(\frac{1}{x})} } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{\log_x(\frac{1}{x})} } \qquad \boxed{ \log_x\left(\frac{1}{x}\right) = \log_x(1) - \log_x(x) } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ \log_x(1) - \log_x(x) } } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ \log_x(1) - \log_x(x^1) } } \\ && \qquad | \quad \log_x(1) =\log_x(x^0)= 0 \\ && \qquad | \quad \log_x(x^1) = 1 \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ 0 - 1 } } \\ &\large{=}& \large{ \log_x(a)+3\times \log_x(a) } \\ &\mathbf{\large{=}}& \mathbf{\large{ 4\times \log_x(a) } }\\ \hline \end{array} \)
+130071 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
