Need a little help getting to the next step (precalculus)?

Question

ok so first i'll give you the question then what i have so far

The electric current I, in amperes (A) in a circuit containing a resistance R, an inductor L, and a voltage source E is given by equation I=E/R (1-e^-Rt/L) where t is time in seconds. Find the time t if I=0.75 amperes, E= 6 volts, R=4.5ohms, and L=2.5H.

ok so i got this so far

I=E/R (1-e^-Rt/L)

0.75=6/4.5(1-e^-4.5t/2.5)

0.75/(6/4.5)=(1-e^-4.5t/2.5)

ok this is where i'm confused??

i was thinking I add or subtract 1 from both side next

I add it seem to look more right then subtracting and getting a negative so i got this now

1.5625= -e^-4.5t/2.5

then i wanted to get rid of 2.5 so i can you the natural log but dont know how to do that enither

cause i want to get

natural log on bother side well hope you can help and understand where i am confused i know what to do just dont know how to get there

Answer 1

You were right up until this:

0.75/(6/4.5)=(1-e^-4.5t/2.5)

I'll simplify the left side and then proceed:

.5625 = 1 - e^-4.5t/2.5

Multiply everything by -1

-.5625 = e^-4.5t/2.5 - 1

Add 1 to both sides

0.4375 = e^-4.5t/2.5

ln 0.4375 = -4.5t/2.5

(-2.5/4.5) (ln 0.4375) = t

The log of a fraction is negative, so you'll end up with a positive value for t.