What is y if t=(1/5)ln3? ...when dy/dt = 5y, y=10 when t=0.?

dy/dt = 5y
(1/y)dy = 5dt
ln(y) = 5t + C₁

When t = 0, then y = 10 :
ln(10) = 5(0) + C₁
Then, ln(y) = 5t + ln(10)

When t = (1/5)ln(3) :
ln(y) = 5*(1/5)ln(3) + ln(10)
ln(y) = ln(3) + ln(10)
ln(y) = ln(30)
y = 30

dy/dt=5y=>
dy/y=5dt=>
ln(y)=5t+C
(0,10)=>C=ln(10)
=>
ln(y)=5t+ln(10)
=>
y=10e^(5t)

When t=ln(3)/5, then
y=10e^[ln(3)]
=>
y=10*3=30