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?

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!

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Answer 1 · 2017-12-17T13:46:04

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 · 2017-12-17T14:14:58

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 $k > 0$

$=>a/kb^x<1$

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

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

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

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

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

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

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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

Explanation:

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The graph y=ab^xy=ab^x passes through (2, 400)(2, 400) and (5,50)(5,50). Find values for aa and bb, and, given that ab^x>kab^x>k for some constant k>0k>0, show that x>log(1600/k)/log2x>log(1600/k)/log2 where log means log to any base?

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

2 Answers

Explanation:

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Impact of this 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!