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# log solving for x Watch

1. Hi i'm trying to solve for x by using log but I'm not sure how to do it

=

I rearranged it to xln3 =

that doesn't look correct, can someone please tell me what I'm doing wrong?

thanks
2. (Original post by HelloGoodbye)
Hi i'm trying to solve for x by using log but I'm not sure how to do it

=

I rearranged it to xln3 =

that doesn't look correct, can someone please tell me what I'm doing wrong?

thanks
Check

That is fine
3. (Original post by TenOfThem)
Check

That is fine
thanks! and I'm trying to do this simultaneous equation

2^x - 3^y = -1
2^(x-1) - 3^(y-1) = -1

would I change the first equation to log like this:

xln2 - yln3 = -1
I'm not sure what to do after that?
4. (Original post by HelloGoodbye)
thanks! and I'm trying to do this simultaneous equation

2^x - 3^y = -1
2^(x-1) - 3^(y-1) = -1

would I change the first equation to log like this:

xln2 - yln3 = -1
I'm not sure what to do after that?
I don't know what you've tried to do there, and it certainly isn't valid!!

I would use the first equation to rewrite 3^y in terms of 2^x (or vice versa) and then substitute into the second equation.
5. (Original post by HelloGoodbye)
thanks! and I'm trying to do this simultaneous equation

2^x - 3^y = -1
2^(x-1) - 3^(y-1) = -1

would I change the first equation to log like this:

xln2 - yln3 = -1
I'm not sure what to do after that?
No

You cannot log individual parts, plus you have not changed the -1

6. (Original post by davros)
I don't know what you've tried to do there, and it certainly isn't valid!!

I would use the first equation to rewrite 3^y in terms of 2^x (or vice versa) and then substitute into the second equation.
ok thanks so I've rearraged the equation to give me 2^x = 3^y-1 and sub'd that into equation 2:

(-1)^-2 - 3^(y-1) + 1=0

- +1=0

Am I doing that right?
7. (Original post by HelloGoodbye)
ok thanks so I've rearraged the equation to give me 2^x = 3^y-1 and sub'd that into equation 2:

(-1)^-2 - 3^(y-1) + 1=0

- +1=0

Am I doing that right?
No

Where has the 1/3^y come from and where has the -2 power come from
8. (Original post by HelloGoodbye)
ok thanks so I've rearraged the equation to give me 2^x = 3^y-1 and sub'd that into equation 2:

(-1)^-2 - 3^(y-1) + 1=0

- +1=0

Am I doing that right?
No, not at all

and

You seem to have complicated things unnecessarily.

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