Udowodnij, że

[alibaba8000]

2/A
Udowodnij, że
e) jeśli a,b,c \(\in\)R-{1} są kolejnymi wyrazami ciągu geometrycznego,
to dla R-{1} prawdziwa jest równość:
\(\frac{log_ax}{log_cx}= \frac{log_ax-log_bx}{log_bx-log_cx}\)

[anka]

\(\frac{log_ax}{log_cx}=\frac{log_ax-log_bx}{log_bx-log_cx}\)

a,b,c - ciąg geometryczny

\(b=aq\)

\(c=aq^2\)

Trzeba udowodnić, że:

\(\frac{log_ax}{log_{aq^2}x}=\frac{log_ax-log_{aq}x}{log_{aq}x-log_{aq^2}x}\)
==================
\(L=\frac{log_ax}{log_{aq^2}x}=\frac{ \frac{logx}{loga}}{ \frac{logx}{log(aq^2)} }=\)

\(\frac{logx}{loga} \cdot \frac{log(aq^2)}{logx}=\frac{log(aq^2)}{loga}\)
==================
\(P=\frac{log_ax-log_{aq}x}{log_{aq}x-log_{aq^2}x}=\)

\(\frac{ \frac{logx}{loga} - \frac{logx}{log(aq)} }{ \frac{logx}{log(aq)} - \frac{logx}{log(aq^2)} }=\)

\(\frac{ \frac{logx \cdot log(aq)-logx \cdot loga}{loga \cdot log(aq)}}{ \frac{logx \cdot log(aq^2)-logx \cdot log(aq)}{log(aq) \cdot log(aq^2)} }=\)

\(\frac{ \frac{logx( log(aq)- loga)}{loga \cdot log(aq)}}{ \frac{logx[log(aq^2)- log(aq)]}{log(aq) \cdot log(aq^2)} }=\)

\(\frac{logx( log(aq)- loga)}{loga \cdot log(aq)} \cdot \frac{log(aq) \cdot log(aq^2)}{logx[log(aq^2)- log(aq)]} =\)

\(\frac{ log(aq)- loga}{loga} \cdot \frac{log(aq^2)}{log(aq^2)- log(aq)} =\)

\(\frac{ log \frac{aq}{a} }{loga} \cdot \frac{log(aq^2)}{log \frac{aq^2}{aq} } =\frac{ log q}{loga} \cdot \frac{log(aq^2)}{log q} =\frac{log(aq^2)}{loga}\)
==================
\(L=P\)