Prove that the graphs of y = logkx are congruent for all positive values of k.?

Question

Prove that the graphs of y = logkx are congruent for all positive values of k.?

Answer 1

What do you mean exactly by congruent? They all start at x = 1/k and go on to positive infinity in x by the constraints on real graphs of logarithms.

Answer 2

y = log(kx) = log k + log x   implies that the plot of y against x is a straight line on a logarithmic x axis ( compare with y = mx  +  c) the  log k is the intercept on the y axis. For different values of k the graph is parallel straight lines with different intercept on y Axis. Hence the graph does not appear to be  congruent.