Solving for unkown exponent

[ChristSaves]

I am try to solve for n in this equation, .726 = (1-1.03^-(n-66))/(1-1.03^-(n-62))
I was going to use logs to solve, but I can't quite figure it out.
I got to this, Log(.726) = Log(1-1.03^-(n-66)) - Log(1-1.03^-(n-62))
Normally, without the (1-) in there, I would move the exponent out and have, Log(.726) = -(n-66)Log(1.03) + (n-62)Log(1.03) but obviously I need to do something with the (1-), though I am not sure what.

Thanks

 

[stapel]

I am try to solve for n in this equation, .726 = (1-1.03^-(n-66))/(1-1.03^-(n-62))
I was going to use logs to solve, but I can't quite figure it out.
I got to this, Log(.726) = Log(1-1.03^-(n-66)) - Log(1-1.03^-(n-62))
Normally, without the (1-) in there, I would move the exponent out and have, Log(.726) = -(n-66)Log(1.03) + (n-62)Log(1.03) but obviously I need to do something with the (1-), though I am not sure what.

Try "cross-multiplying" and using exponent properties:

. . . . .\(\displaystyle 0.726\left(1\, -\, 1.03^{62-n}\right)\, =\, 1\, -\, 1.03^{66-n}\)

. . . . .\(\displaystyle 0.726\, -\, 0.726\times 1.03^{62-n}\, =\, 1\, -\, 1.03^{4+62-n}\)

. . . . .\(\displaystyle 1.03^{4+62-n}\, -\, 0.725\times 1.03^{62-n}\, =\, 1\, -\, 0.726\, =\, 0.274\)

Factor, simplify, and then apply logs to the result. :wink: