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# Solving equation using logarithms? watch

1. 3^t+1 = 6 + 3^2t-1

I just guessed on a calculator and it was 1.
2. Just to clarify, is your equation:

?

If it is then note:

as well as: and

Can you let let:

then re-write your original equation to reveal a familiar type of equation?
3. (Original post by simon0)
Just to clarify, is your equation:

?
Yes
4. Nvm. Much more complicated method which I said and wont work always.

do what poster above said.
5. Just to clarify, I had re-written my post above to include a method.
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
7. (Original post by Sakura-Sama)
ln [3^(t+1)] - ln [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.
8. (Original post by Sakura-Sama)
ln [3^(t+1)] - ln [3^(2t-1)] = 6
ln [3^(t+1) / 3^(2t-1)] =6
Note, generally:

where .
9. I tried working this out as well but I can't seem to get to the solution?
10. it is a disguised quadratic... but you need an extra step than in typical examples. you need to split the 3{2t -1} so the 3{2t } appears without the -1
11. (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?
12. I got the quadratic equation 1/3y^2 + 6 -3y=0
is this right?
13. (Original post by user20167)
I got the quadratic equation 1/3y^2 + 6 -3y=0
is this right?
Looks good, can you finish it off?
14. (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
15. (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"?
16. (Original post by simon0)
Remember your substitution. What was "y"?
1.63 and 1 are the solutions
17. (Original post by user20167)
1.63 and 1 are the solutions
Looks good.

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