Find a formula of exponential growth or decay from two points - High school Algebra I Exercises

f\left(1\right)=3 and f\left(2\right)=8.

Assuming that f grows exponentially, what is the formula of growth of f ?

f\left(x\right)=8\cdot \left( \dfrac{5}{3} \right)^{x}

f\left(x\right)=\dfrac{9}{8}\cdot \left( \dfrac{5}{3} \right)^{x}

f\left(x\right)=\dfrac{9}{8}\cdot \left( \dfrac{8}{3} \right)^{x}

f\left(x\right)=8\cdot \left( \dfrac{3}{8} \right)^{x}

The general equation for an exponential function is:

f\left(x\right)=a\left(b\right)^{x}

Here:

f\left(1\right)=3 is equivalent to a\left(b\right)^{1}=3.
f\left(2\right)=8 is equivalent to a\left(b\right)^{2}=8.

Dividing the two equations, we get:

\dfrac{a\left(b\right)^{1}}{a\left(b\right)^{2}}=\dfrac{3}{8}

Which is equivalent to:

\dfrac{1}{b}=\dfrac{3}{8}

So, we conclude that:

b=\dfrac{8}{3}

That means:

a\cdot\dfrac{8}{3}=3

Which solved for a :

a=\dfrac{9}{8}

We conclude that:

f\left(x\right)=\dfrac{9}{8}\cdot \left( \dfrac{8}{3} \right)^{x}

f\left(1\right)=5 and f\left(2\right)=15.

Assuming that f grows exponentially, what is the formula of growth of f ?

f\left(x\right)=15\cdot \left( 3 \right)^{x-1}

f\left(x\right)=\dfrac{1}{3}\cdot \left( 5 \right)^{x}

f\left(x\right)=15\cdot \left( \dfrac{1}{3} \right)^{x}

f\left(x\right)=\dfrac{5}{3}\cdot \left( 3 \right)^{x}

The general equation for an exponential function is:

f\left(x\right)=a\left(b\right)^{x}

Here:

f\left(1\right)=5 is equivalent to a\left(b\right)^{1}=5.
f\left(2\right)=15 is equivalent to a\left(b\right)^{2}=15.

Dividing the two equations, we get:

\dfrac{a\left(b\right)^{1}}{a\left(b\right)^{2}}=\dfrac{5}{15}

Which is equivalent to:

\dfrac{1}{b}=\dfrac{1}{3}

So, we conclude that:

b=3

That means:

a\cdot3=5

Which solved for a :

a=\dfrac{5}{3}

We conclude that:

f\left(x\right)=\dfrac{5}{3}\cdot \left(3 \right)^{x}

f\left(1\right)=100 and f\left(3\right)=25.

Assuming that f decays exponentially, what is the formula of decay of f ?

f\left(x\right)=100\cdot \left( \dfrac{1}{4} \right)^{x}

f\left(x\right)=50\cdot \left( \dfrac{1}{2} \right)^{x-1}

f\left(x\right)=100\cdot \left( \dfrac{1}{2} \right)^{x}

f\left(x\right)=200\cdot \left( \dfrac{1}{2} \right)^{x}

The general equation for an exponential function is:

f\left(x\right)=a\left(b\right)^{x}

Here:

f\left(1\right)=100 is equivalent to a\left(b\right)^{1}=100.
f\left(3\right)=25 is equivalent to a\left(b\right)^{3}=25.

Dividing the two equations, we get:

\dfrac{a\left(b\right)^{1}}{a\left(b\right)^{3}}=\dfrac{100}{25}

Which is equivalent to:

\dfrac{1}{b^{2}}=4

So, we conclude that:

b=\dfrac{1}{2}

That means:

a\cdot\dfrac{1}{2}=100

Which solved for a :

a=200

We conclude that:

f\left(x\right)=200\cdot \left(\dfrac{1}{2} \right)^{x}

f\left(-1\right)=\dfrac{1}{4} and f\left(2\right)=2.

Assuming that f grows exponentially, what is the formula of growth of f ?

f\left(x\right)=\dfrac{1}{2}\cdot \left( 2 \right)^{x}

f\left(x\right)=4\cdot \left( \dfrac{1}{8} \right)^{x-1}

f\left(x\right)=\dfrac{1}{4}\cdot \left( 2 \right)^{x}

f\left(x\right)=\dfrac{1}{8}\cdot \left( 2 \right)^{x}

The general equation for an exponential function is:

f\left(x\right)=a\left(b\right)^{x}

Here:

f\left(-1\right)=\dfrac{1}{4} is equivalent to a\left(b\right)^{-1}=\dfrac{1}{4}.
f\left(2\right)=2 is equivalent to a\left(b\right)^{2}=2.

Dividing the two equations, we get:

\dfrac{a\left(b\right)^{-1}}{a\left(b\right)^{2}}=\dfrac{\dfrac{1}{4}}{2}

Which is equivalent to:

\dfrac{1}{b^{3}}=\dfrac{1}{8}

So, we conclude that:

b=2

That means:

a\cdot2^{-1}=\dfrac{1}{4}

Which solved for a :

a=\dfrac{1}{2}

We conclude that:

f\left(x\right)=\dfrac{1}{2}\cdot \left( 2 \right)^{x}

f\left(1\right)=3 and f\left(4\right)=24.

Assuming that f grows exponentially, what is the formula of growth of f ?

f\left(x\right)=\dfrac{3}{2}\cdot \left( 2 \right)^{x}

f\left(x\right)=3\cdot \left( 2 \right)^{x-1}

f\left(x\right)=\dfrac{2}{3}\cdot \left( 2 \right)^{x}

f\left(x\right)=\dfrac{3}{4}\cdot \left( 2 \right)^{x}

The general equation for an exponential function is:

f\left(x\right)=a\left(b\right)^{x}

Here:

f\left(1\right)=3 is equivalent to a\left(b\right)^{1}=3.
f\left(4\right)=24 is equivalent to a\left(b\right)^{4}=24.

Dividing the two equations, we get:

\dfrac{a\left(b\right)^{1}}{a\left(b\right)^{4}}=\dfrac{3}{24}

Which is equivalent to:

\dfrac{1}{b^{3}}=\dfrac{1}{8}

So, we conclude that:

b=2

That means:

a\cdot2=3

Which solved for a :

a=\dfrac{3}{2}

We conclude that:

f\left(x\right)=\dfrac{3}{2}\cdot \left( 2 \right)^{x}

f\left(-3\right)=80 and f\left(0\right)=10.

Assuming that f decays exponentially, what is the formula of decay of f ?

f\left(x\right)=\dfrac{1}{8}\cdot \left( 2 \right)^{x}

f\left(x\right)=10\cdot \left( \dfrac{1}{2} \right)^{x}

f\left(x\right)=80\cdot \left( \dfrac{1}{8} \right)^{x-1}

f\left(x\right)=80\cdot \left( \dfrac{1}{8} \right)^{x+1}

The general equation for an exponential function is:

f\left(x\right)=a\left(b\right)^{x}

Here:

f\left(-3\right)=80 is equivalent to a\left(b\right)^{-3}=80.
f\left(0\right)=10 is equivalent to a\left(b\right)^{0}=10.

Dividing the two equations, we get:

\dfrac{a\left(b\right)^{-3}}{a\left(b\right)^{0}}=\dfrac{80}{10}

Which is equivalent to:

\dfrac{1}{b^{3}}=8

So, we conclude that:

b=\dfrac{1}{2}

That means:

a\cdot \left( \dfrac{1}{2}\right)^{0}=10

Which solved for a :

a=10

We conclude that:

f\left(x\right)=10\cdot \left( \dfrac{1}{2} \right)^{x}

f\left(-1\right)=5 and f\left(1\right)=45.

Assuming that f grows exponentially, what is the formula of growth of f ?

f\left(x\right)=15\cdot 3^{x}

f\left(x\right)=15\cdot \left( \dfrac{1}{3} \right)^{x}

f\left(x\right)=5\cdot 9^{x-1}

f\left(x\right)=5\cdot 3^{x}

The general equation for an exponential function is:

f\left(x\right)=a\left(b\right)^{x}

Here:

f\left(-1\right)=5 is equivalent to a\left(b\right)^{-1}=5.
f\left(1\right)=45 is equivalent to a\left(b\right)^{1}=45.

Dividing the two equations, we get:

\dfrac{a\left(b\right)^{-1}}{a\left(b\right)^{1}}=\dfrac{5}{45}

Which is equivalent to:

\dfrac{1}{b^{2}}=\dfrac{1}{9}

So, we conclude that:

b=3

That means:

a\cdot 3^{1}=45

Which solved for a :

a=15

We conclude that:

f\left(x\right)=15\cdot 3^{x}