What is the inverse of #y=3log(5x)-ln(5x^3)#? ?

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Answer 1 · 2016-06-22T14:36:50

#y = 1.33274 xx10^((-0.767704 x)/3)# for #0 < x < oo#

Supposing that #log a = log_{10}a, ln a = log_e a#

For #0 < x < oo#

#y = log_e(5x)^3/log_e 10-log_e(5x)^3+log_e 25#

#y log_e10 = (1-log_e10)log_e(5x)^3+log_e25 xxlog_e 10#

#log_e(5x)^3=(y log_e10 - log_e25 xxlog_e 10)/ (1-log_e10)#

#(5x)^3=c_0e^{c_1y}#

where #c_0 = e^(-(log_e25 xxlog_e 10)/ (1-log_e10))#
and #c_1 = log_e10/(1-log_e10)#

Finally

#x = 1/5 c_0^{1/3} xx e^{c_1/3 y}#

or

#x = 1.33274 xx10^((-0.767704 y)/3)#

Red #y=3log(5x)-ln(5x^3)#
Blue #y = 1.33274 xx10^((-0.767704 x)/3)#

enter image source here