Basic log question, completely lost

[Chadlee88]

Basic log question, completely lost! :(

Homework Statement

Prove that y = a(1+b)^t is equivalent to ln y = a+bt

Homework Equations

The Attempt at a Solution

i'm really confused with how they could possibly be equal.

y = a(1+b)^t
t = ln y/ln a(1+b)
ln y = t x ln (a+b)
ln y = ln (a+b)^t

Could someone please help.

 

[neutrino]

Of course they are not equivalent for all values of t. Are you sure you have the question right?

And, BTW, steps 3 and 4 of your attempt doesn't seem right, although it doesn't matter now.

 

[Chadlee88]

ohh sorry, i made a mistake.

This is the original question:

prove that yt = y0(1+r)^t is equivalent to ln yt = y0 + rt

i just made y0 = a and r = b previously

My Attempt:
yt = y0(1+r)^t
t = ln yt/ln y0(1+r)

you said i was wrong previous, where do i go from this step then?? Tanx a heap 4 helping.

 

[HallsofIvy]

ohh sorry, i made a mistake.

This is the original question:

prove that yt = y0(1+r)^t is equivalent to ln yt = y0 + rt

i just made y0 = a and r = b previously

My Attempt:
yt = y0(1+r)^t
t = ln yt/ln y0(1+r)

you said i was wrong previous, where do i go from this step then?? Tanx a heap 4 helping.

First thing you should do is go back and look at the problem again! What you have stated still is not true. From yt= yo(1+r)^t the best you can say, by taking ln of both sides, is ln(yt)= ln(y0)+ t ln(1+r). In general that is NOT the same as "ln yt= y0+ rt".

(You can say that "to first order" (ignoring higher powers of t) yt (NOT ln(yt)) is approximately equal to y0+ rt.)