How do you solve #5(10^y)=1+2(10^y)#?

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Answer 1 · 2018-05-10T17:33:14

I tried this:

Let us write it as:

#5(10^y)-2(10^y)=1#

collect:

#(10^y)[5-2]=1#

#(10^y)*3=1#

#10^y=1/3#

take the log in base #10# of both sides:

#log_(10)(10^y)=log_(10)(1/3)#

giving:

#y==log_(10)(1/3)=-0.477121#