SOLUTION: How would you go about solving these problems 2e^(x+1)-7=3 5log32^(x+1)=3 (note: 32 is the base)

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Question 34223: How would you go about solving these problems
2e^(x+1)-7=3

5log32^(x+1)=3 (note: 32 is the base)
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
How would you go about solving these problems
2e^(x+1)-7=3
E^(X+1)^2=3+7=10...TAKING LOGS
(X+1)^2*(LOG(E))=LOG(10)=1
(X+1)^2=10/LOG(E)
X+1=SQRT{10/LOG(E)}
X=SQRT{10/LOG(E)}-1=1.08

5log32^(x+1)=3 (note: 32 is the base)
LOG{(X+1)^5}TO BASE 32=3
(X+1)^5=32^3=(2^5)^3=(2^3)^5
X+1=2^3=8
X=8-1=7