Write (equation) as a single logarithm.

Attached is the equation and the solution.

Did I solve it correctly?

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- Dec 17th 2008, 06:12 AMmwokWrite as single logarithm
Write (equation) as a single logarithm.

Attached is the equation and the solution.

Did I solve it correctly? - Dec 17th 2008, 06:54 AMiLikeMaths
no you did not solve it correctly, remember the rules of log,

is the same as

and is the same as

also

+ is the same as

- Dec 17th 2008, 06:57 AMmwok
Ah. Is this correct?

- Dec 17th 2008, 07:02 AMiLikeMaths
and there it is, as a single logarithm, u can try to put any number back into the equation to check your working e.g. let x = 5

if u put this with x equals 5 into your calculator you will get 5.999 or something

and if you put x=5 into the single logarithm

you will get the same 5.99 so you know its correct

hope you understand - Dec 17th 2008, 07:04 AMmwok
Thanks!

- Dec 17th 2008, 07:04 AMMoo
- Dec 17th 2008, 07:15 AMmwok
Wow.

Okay, so the final solution is attached.

EDIT:

Why does the "x" come out of ln(x)^2?? - Dec 17th 2008, 07:18 AMiLikeMaths
because of the two infront of it, note that this only works with logs/ln

- Dec 17th 2008, 07:20 AMmwok
- Dec 17th 2008, 07:20 AMMoo
- Dec 17th 2008, 07:21 AMMoo
- Dec 17th 2008, 07:26 AMmwok
Ah I see how the exponents work now BUT why does x^2 come out of the ln(x^2)? What happens to the ln?

- Dec 17th 2008, 07:28 AMMoo
- Dec 17th 2008, 07:29 AMmwok
Thanks!

- Dec 17th 2008, 07:33 AMiLikeMaths
some times != means does not equal, well thats what i thought he/she meant