Logarithm Question

whistlin_boy

I have missed maths for the last week because of music practice for de open day (ughh) and was told on friday have to do all de work I missed, including this revison question on logarithms

5^2n-1=2^n

It is probably something really easy, but I cant make heads or tails of it:o

Thanks!

felic

These types of questions are fine once you know what to do. You want to use In

so something like this:

5^2n-1=2^n

In5^2n-1= In2^n (manipulating both sides by the same In)
(2n-1)In5 = nIn2 (look up rules of logs...)
2nIn5 - In5 = nIn2 (multiply them out)
2nIn5 - nIn2 = In5 (bring the n's over to the one side)
n(2In5 - In2) = In5 (factorize n)
n = In5 / (2In5 - In2)

Then you can get an answer for that from your calculator... or leave it like this)

whistlin_boy

felic wrote: »

These types of questions are fine once you know what to do. You want to use In

so something like this:

5^2n-1=2^n

In5^2n-1= In2^n (manipulating both sides by the same In)
(2n-1)In5 = nIn2 (look up rules of logs...)
2nIn5 - In5 = nIn2 (multiply them out)
2nIn5 - nIn2 = In5 (bring the n's over to the one side)
n(2In5 - In2) = In5 (factorize n)
n = In5 / (2In5 - In2)

Then you can get an answer for that from your calculator... or leave it like this)

Thanks....usually Im good at maths but this question just confused me:o