Logs

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

Reply 1

TenOfThem

18

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

2^{x+1} = 3^{3x-2}

OP

yes

Can you put that in brackets please.

Reply 4

Rascacielos

19

I thought this thread was entitled "legs." What am I like?

Reply 5

Bloxorus

16

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.

Posted from TSR Mobile

Reply 6

TenOfThem

18

Original post by jacksonmeg

yes

Are you aware that

a^x = b^y

can be re-written as

y log a = x log b

Reply 7

jacksonmeg

OP

11

Yes, but I don't know how to solve them, need help urgently, can't find any worked examples like these

(edited 10 years ago)

Reply 8

joostan

13

Original post by jacksonmeg

Yes, but I don't know how to solve them, need help urgentl

What have you done so far? :smile:

Reply 9

jacksonmeg

OP

11

(X+1)log2 - (3x-2)log3 = 0 :/

I know when you minus you divide them but I got the wrong answer :/

(edited 10 years ago)

Reply 10

Dingusnin

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 :smile:

Reply 11

joostan

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 :smile:

Perhaps spoiler your solution so the OP can have a go. :smile:

Full solutions should only be provided if absolutely necessary.
Please Read.

Reply 12

jacksonmeg

OP

11

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 :smile:

i don't get what you've done :/