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# Logs Watch

1. could anyone tell me how to solve logs like these

Find x

1.) 2^x+1 = 3^3x-2

2.) 5^2x-3 = 7^3x-1

I don't have a clue :P
2. (Original post by jacksonmeg)
could anyone tell me how to solve logs like these

Find x

1.) 2^x+1 = 3^3x-2

2.) 5^2x-3 = 7^3x-1

I don't have a clue :P
Is the first question
3. yes
4. Can you put that in brackets please.
5. I thought this thread was entitled "legs." What am I like?
6. Take logs of both sides to the base of either the 3 or 2 for the first one so the log cancels out on one of the sides. Then just work from there.

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7. (Original post by jacksonmeg)
yes
Are you aware that can be re-written as
8. Yes, but I don't know how to solve them, need help urgently, can't find any worked examples like these
9. (Original post by jacksonmeg)
Yes, but I don't know how to solve them, need help urgentl
What have you done so far?
10. (X+1)log2 - (3x-2)log3 = 0 :/

I know when you minus you divide them but I got the wrong answer :/
11. (Original post by jacksonmeg)
could anyone tell me how to solve logs like these

Find x

1.) 2^x+1 = 3^3x-2

2.) 5^2x-3 = 7^3x-1

I don't have a clue :P

You need to take logs of both sides (Any base, but either 2 or 3 would be best). Then one you have that, use log laws to take out the powers:
x+1 = (3x-2)log(2,3)
x+1 = 3x*log(2,3) -2log(2,3)
x+3x*log(2,3) = -1-2log(2,3)
x(1+3*log(2,3)) = -1-2log(2,3)
x = (-1-2log(2,3))/(1+3log(2,3))

There we go
12. (Original post by Dingusnin)
You need to take logs of both sides (Any base, but either 2 or 3 would be best). Then one you have that, use log laws to take out the powers:
x+1 = (3x-2)log(2,3)
x+1 = 3x*log(2,3) -2log(2,3)
x+3x*log(2,3) = -1-2log(2,3)
x(1+3*log(2,3)) = -1-2log(2,3)
x = (-1-2log(2,3))/(1+3log(2,3))

There we go
Perhaps spoiler your solution so the OP can have a go.
Full solutions should only be provided if absolutely necessary.
13. (Original post by Dingusnin)
You need to take logs of both sides (Any base, but either 2 or 3 would be best). Then one you have that, use log laws to take out the powers:
x+1 = (3x-2)log(2,3)
x+1 = 3x*log(2,3) -2log(2,3)
x+3x*log(2,3) = -1-2log(2,3)
x(1+3*log(2,3)) = -1-2log(2,3)
x = (-1-2log(2,3))/(1+3log(2,3))

There we go
i don't get what you've done :/
14. (Original post by jacksonmeg)
i don't get what you've done :/
Take logs of both sides.
E.g
Spoiler:
Show

How do you think we should proceed?
15. Sorry, i will spoiler-it next time

what i have done is that i have taken logs (of base 2) on either side, and rearranged in order to solve for x.
16. What's the comma mean? Logs can go die, going to fail C2
17. (Original post by jacksonmeg)
What's the comma mean? Logs can go die, going to fail C2
Sorry, i should have posted using LaTeX, What i meant by that is:
log(base,x)
in other words: log(2,8) = 3, Because 2^3 = 8
18. (Original post by joostan)
Take logs of both sides.
E.g
Spoiler:
Show

How do you think we should proceed?
I don't have a clue :/
19. (Original post by jacksonmeg)
I don't have a clue :/
Well how would you go about isolating x?
20. Idk :/ none of the rules I've got seem to work

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