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[Tulac]

Ok I have huge problems solving these math tasks so if someone could me me out I would appreciate it very much...
Please if you know how to solve it, post the whole proccess of solving it, not just the result...
(the tasks are logarythms and exponential equation)

[Tulac]

*Bumps*
Come on people someone must know the answers...

[BeefontheBone]

Solving for x, right? Gimme a sec.

For the first one I get x=log(2+-root3) - the log in base 10 of 2 plus or minus the square root of 3.

First off, take the 3 on the left hand side (LHS) inside the brackets and expand the square on the RHS to get:

3*10^(2x) + 6*10^x = 4*10^(2x) + 2*10^x + 1

then tidy that up and move everything to the RHs to get a quadratic equation in 10^x:

10^(2x) - 4*10^x + 1 =0

Then that has real solutions where 10^x = (4+-root12)/2 = 2+-root3

so x=log(2+-root3)

[Tulac]

Originally posted by BeefontheBone@Jun 5 2005, 09:19 PM
Solving for x, right?

Yes!

[Sebatianos]

OK - second example is easy: X=0

Just put it all together and you get 17X = 65X (and the only way for that to be true x = 0.

I can't help you with no. 3, because I completely forgoten how to do logatihms.

Give me a moment for the 1st one.

[Tulac]

Meh I always ended up with some weird errm mi to kažemo razlomci, not sure how to call it in English, I always got something like 12/5^x=12...

[BeefontheBone]

(12/5)^x = 12 is right f0r that, so x=log[12/5]12 - x is the log of 12 in base 12/5. that can probably be simplified a bit better, but a calculator will work it out - I had about 4 hours' sleep this morning...

On the third one, does it mean that the first log term is squared but they're both in base 2? In fact, that's not an equation, you sure there shouldn't be another bit in there? Is it that term =0?

[Tulac]

Originally posted by BeefontheBone@Jun 5 2005, 09:19 PM
so x=log(2+-root3)

Meh I've got somewhere in the lines of that but I messed inside the quadratic equation, I forgot to put minus in front of b...

[Tulac]

Originally posted by BeefontheBone@Jun 5 2005, 09:37 PM
(12/5)^x = 12 is right f0r that, so x=log[12/5]12 - x is the log of 12 in base 12/5. that can probably be simplified a bit better, but a calculator will work it out - I had about 4 hours' sleep this morning...

I don't need to get a precise result I think that's enough, hmmph it seems I got that one too...
I guess I need more confidence about my math skills, now what about the third one?

Sorry for double post people...

Can't do your own homework, eh? I know this would be closed right away if it was posted in the forums where I moderate.