e^2x-1 = 5 + k

[Chris is awesome]

e^(2x-1) = 5 + k

Assuming all of the LHS is the power of e.

 

You can take Ln of all terms.

 

Ln(e^2x-1) = Ln5 + k (assuming k is a constant Ln(k) is still k so there is no need to write it)

 

2x-1 = Ln5 + k (canceling Ln and e)

 

2x = Ln5 + k + 1( + 1 to both sides)

 

x = (Ln5 + k +1)/2 ( divide by 2)

 

Probably wrong though.

[ShadowCaptain]

7

that is all

[Snapy1]

Ahhhhh im only in Algebra 2 *goes in the corner and crys*

[DIV1D3]

I was in Extend. Maths C in Yr12 and I have no clue what is happening on this thread. I feel as if the school I went to had the worst teaching, we'd revisit shite we already knew that we already did in Yr6 in Yr10!

[TideRower]

Ln(e^2x-1) = Ln5 + k (assuming k is a constant Ln(k) is still k so there is no need to write it)

That's not really correct, because you can't use the same constant for two different uses. You could however write ln(5+k)=k₀   (k,k₀ are constants).

 

Ahhhhh im only in Algebra 2 *goes in the corner and crys*

 

  I'm also in Algebra 2, but at a higher level. It's just funny that they call it the same.