hepl please

[Spyros13]

i have this : 15,76=0.43*(20)ⁿ

i have to calculate n...

15.76/0.43=20ⁿ

36.65=20ⁿ

n=log₂₀36.65

is this right???

 

[hypervalent_iodine]

No, it isn't right.

 

Remember this rule:

 

log a^b = b * log a

[Spyros13]

but is this steps right?

36.65=20ⁿ =>

=>n=log₂₀36.65

Edited February 12, 2011 by Spyros13

[Blahah]

but is this steps right?

36.65=20ⁿ =>

=>n=log₂₀36.65

No, this step is incorrect.

 

you have

36.65=20ⁿ

first put the unknown on the left:

20ⁿ = 36.65

to get rid of the exponent you can take the log of 20ⁿ. Make sure you do the same to both sides.

n log20 = log36.65

then you should be able to find n.

Edited February 12, 2011 by Blahah

[Shadow]

hypervalent_iodine and Blahah, you are both incorrect. Spyros, you're answer is correct; the guys above didn't notice you took the base 20 logarithm of both sides, not the natural/base 10 log:

 

[math]36.65 = 20^n[/math]

 

[math]\log_{20}(36.65) = \log_{20}(20^n)[/math]

 

[math]\log_{20}(36.65) = n \cdot \log_{20}(20)[/math]

 

[math]n = \log_{20}(36.65) \approx 1.202181[/math]

Edited February 12, 2011 by Shadow

[Blahah]

Whoops sorry - Shadow's right, I didn't notice that yours was log base 20. However, I think my solution is still correct as well.

 

eidt: ugh, I also formatted the text wrongly in my answer. Corrected now. Maybe I should wait until I'm better at maths before advising others!

Edited February 12, 2011 by Blahah

[hypervalent_iodine]

and this is why I'm a chemist.