Question

The graph `y=ab^x` passes through `(2, 400)` and `(5,50)`. Find values for `a` and `b`, and, given that `ab^x>k` for some constant `k>0`, show that `x>log(1600/k)/log2` where log means log to any base?

I am very confident that `b=1/2` and `a=1600` but I'm not sure how to go about this proof. Thanks!

Answer 1

Please see below.

Explanation:

As `y=ab^x` and it passes though `(2,400)` and `(5,50)`, we have

`400=ab^2` and `50=ab^5`

Dividing latter by former, we get `b^3=1/8` or `b=1/2`

As such `400=axx(1/2)^2` or `400=axx1/4`

and hence `a=400xx4=1600`

Let for some `x`, we have `ab^x>k`, where `k>0`

then `log_n1600+xlog_n(1/2)>log_nk`

or `xlog_n(1/2)>log_nk-log_n1600`

or `0-xlog_n2> log_nk-log_n1600`

or `xlog_n2> log_n1600-log_nk`

i.e. `xlog_n2> log_n(1600/k)`

or `x > log_n(1600/k)/log_n2`

Answer 2

Given `y=ab^x` passes through `(2,400) and (5,50)`

So `400=ab^2.....(1)`

and

`50=ab^5.....(2)`

Dividing (2) by (1) we get

`b^3=50/400=1/8=(1/2)^3`

`=>b=1/2`

Inserting the value of b in (1) we get

`a*(1/2)^2=400`

`=>a=1600`

Now

`ab^x < k` where `k > 0`

`=>a/kb^x<1`

`=>a/k < 1/b^x`

`=>a/k< (1/b)^x`

`=>1600/k<(2)^x`

`=>log(1600/k)< log(2)^x`

`=>log(1600/k)< xlog2`

`=>x>(log(1600/k))/log2`