Solving equation using logarithms?
3^t+1 = 6 + 3^2t-1
I just guessed on a calculator and it was 1.
The answer is, how?
I just guessed on a calculator and it was 1.
The answer is, how?
Original post by simon0
Just to clarify, is your equation:
?
?
Yes
Nvm. Much more complicated method which I said and wont work always.
do what poster above said.
do what poster above said.
(edited 6 years ago)
Original post by y.u.mad.bro?
Start off by taking log of both side. What do you get?
ln [3^(t+1)] - ln [3^(2t-1)] = 6
ln [3^(t+1) / 3^(2t-1)] =6
Original post by Sakura-Sama
ln [3^(t+1)] - ln [3^(2t-1)] = 6
ln [3^(t+1) / 3^(2t-1)] =6
ln [3^(t+1) / 3^(2t-1)] =6
Sorry, I thought it wa easier to use logs but it actually isn't. You are better off realising that it is a disguised quadratic.
Original post by Sakura-Sama
ln [3^(t+1)] - ln [3^(2t-1)] = 6
ln [3^(t+1) / 3^(2t-1)] =6
ln [3^(t+1) / 3^(2t-1)] =6
Note, generally:
where .
(edited 6 years ago)
I tried working this out as well but I can't seem to get to the solution? 
Original post by user20167
I tried working this out as well but I can't seem to get to the solution?
What if I stated:
and
Does this help?
I got the quadratic equation 1/3y^2 + 6 -3y=0
is this right?
is this right?
Original post by user20167
I got the quadratic equation 1/3y^2 + 6 -3y=0
is this right?
is this right?
Looks good, can you finish it off?
Original post by simon0
Looks good, can you finish it off?
I get the solutions 3 and 6 which don't work when put back into the equation though
Original post by user20167
I get the solutions 3 and 6 which don't work when put back into the equation though
Remember your substitution. What was "y"?
Original post by simon0
Remember your substitution. What was "y"?
1.63 and 1 are the solutions
Original post by user20167
1.63 and 1 are the solutions
Looks good.
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