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!

2 Answers
Dec 17, 2017

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#

Dec 17, 2017

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#