SOLUTION: Please help me solve these equations 4^3x+1=28 ln(x+1)=3 4log(base 2x)+log(base 2)405

Question 250973: Please help me solve these equations
4^3x+1=28
ln(x+1)=3
4log(base 2x)+log(base 2)405

Answer by dabanfield(803) (Show Source): You can put this solution on YOUR website!
(1.) 4^3x+1=28
(2.) ln(x+1)=3
(3.) 4log(base 2x)+log(base 2)405
(1.) Take the log of both sides:
log (4^(3x+1)) = log 28
(3x+1)* log 4 = log 28
3x+1 = log 28/log 4
3x = (log 28/log 4) - 1
x = [(log 28/log 4 - 1)]/3
(2.) Since ln (x+1) = 3 we know then by definition that:
e^3 = x+1
x = e^3 - 1
(3.) Assuming we have 4*log (base 2) x = log (base 2) 405
then
log (base 2) x^4 = log (base 2) 405
so:
x^4 = 405
x = 405^(1/4)

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