# 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 $b = \frac{1}{2}$ and $a = 1600$ but I'm not sure how to go about this proof. Thanks!

Dec 17, 2017

#### Explanation:

As $y = a {b}^{x}$ and it passes though $\left(2 , 400\right)$ and $\left(5 , 50\right)$, we have

$400 = a {b}^{2}$ and $50 = a {b}^{5}$

Dividing latter by former, we get ${b}^{3} = \frac{1}{8}$ or $b = \frac{1}{2}$

As such $400 = a \times {\left(\frac{1}{2}\right)}^{2}$ or $400 = a \times \frac{1}{4}$

and hence $a = 400 \times 4 = 1600$

Let for some $x$, we have $a {b}^{x} > k$, where $k > 0$

then ${\log}_{n} 1600 + x {\log}_{n} \left(\frac{1}{2}\right) > {\log}_{n} k$

or $x {\log}_{n} \left(\frac{1}{2}\right) > {\log}_{n} k - {\log}_{n} 1600$

or $0 - x {\log}_{n} 2 > {\log}_{n} k - {\log}_{n} 1600$

or $x {\log}_{n} 2 > {\log}_{n} 1600 - {\log}_{n} k$

i.e. $x {\log}_{n} 2 > {\log}_{n} \left(\frac{1600}{k}\right)$

or $x > {\log}_{n} \frac{\frac{1600}{k}}{\log} _ n 2$

Dec 17, 2017

Given $y = a {b}^{x}$ passes through $\left(2 , 400\right) \mathmr{and} \left(5 , 50\right)$

So $400 = a {b}^{2.} \ldots . \left(1\right)$

and

$50 = a {b}^{5.} \ldots . \left(2\right)$

Dividing (2) by (1) we get

${b}^{3} = \frac{50}{400} = \frac{1}{8} = {\left(\frac{1}{2}\right)}^{3}$

$\implies b = \frac{1}{2}$

Inserting the value of b in (1) we get

$a \cdot {\left(\frac{1}{2}\right)}^{2} = 400$

$\implies a = 1600$

Now

$a {b}^{x} < k$ where $k > 0$

$\implies \frac{a}{k} {b}^{x} < 1$

$\implies \frac{a}{k} < \frac{1}{b} ^ x$

$\implies \frac{a}{k} < {\left(\frac{1}{b}\right)}^{x}$

$\implies \frac{1600}{k} < {\left(2\right)}^{x}$

$\implies \log \left(\frac{1600}{k}\right) < \log {\left(2\right)}^{x}$

$\implies \log \left(\frac{1600}{k}\right) < x \log 2$

$\implies x > \frac{\log \left(\frac{1600}{k}\right)}{\log} 2$