Given #W=3^(x+y)#, how do you solve for x?

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Answer 1 · 2016-11-08T01:01:27

#lnW = ln(3^(x + y))#

#lnW = (x + y)ln3#

#lnW = xln3 + yln3#

#lnW - yln3 = xln3#

#(lnW - yln3)/ln3 = x#

#(ln(W/3))/ln3 = x#

Hopefully this helps!

Answer 2 · 2016-11-08T01:29:14

#x=log_3 W-y=log W/log 3 - y#.

Inversely,

#x+y=log_3W#

#x=log_3 W-y#.