If
#color(white)("XXX")1€=0.73£#
then
#color(white)("XXX")1/(0.73)€=1£# and since #1/(0.73) ~~1.37#
#rarrcolor(white)("XXX") k£~~(1.37 * k) €#
#rarrcolor(white)("XXX") 50£~~(1.37 * 50) € =68.5 €#
then, for verification, using the original function:
#color(white)("XXX")68.5 €=(0.73 * 68.5)£=50.005£# (for a reasonably accurate equivalence)
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I am quite unhappy using the #€# and #£# symbols as both variable amounts and unit designates. However, I have provided this answer in the form it was asked.
