Asked by Anonymous
Solve each equation for x
e^(4x+1)=p
log (kx)/ log (5) = c
e^(4x+1)=p
log (kx)/ log (5) = c
Answers
Answered by
Reiny
e^(4x+1)=p
ln(e^(4x+1)) = ln p
(4x+1) lne = lnp , but lne =1
4x+1 = lnp
4x = lnp - 1
x = (lnp - 1)/4
log (kx)/ log (5) = c
log( kx/log5) = c
kx/log5 = 10^c
kx = log5(10^c)
x = log5(10^c)/k
ln(e^(4x+1)) = ln p
(4x+1) lne = lnp , but lne =1
4x+1 = lnp
4x = lnp - 1
x = (lnp - 1)/4
log (kx)/ log (5) = c
log( kx/log5) = c
kx/log5 = 10^c
kx = log5(10^c)
x = log5(10^c)/k
Answered by
Reiny
scrap the solution to the second equation, I misread the question
here is the correct solution:
log (kx)/ log (5) = c
log kx = c log5
kx = 10^(c log5)
x = 10^(c log5) / k
here is the correct solution:
log (kx)/ log (5) = c
log kx = c log5
kx = 10^(c log5)
x = 10^(c log5) / k
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