Question

Hi there! Can someone help me solve these equations? Thanks!

Part A: Explain why the x-coordinates of the points where the graphs of the equations`y = 2^x` and `y = 4^(x−2)` intersect are the solutions of the equation `2^x = 4^(x−2)`.

Part B: Make tables to find the solution to `2^x = 4^(x−2)`. Take the integer values of `x` between `−4` and `4`.

Part C: How can you solve the equation `2^x = 4^(x−2)` graphically?

Answer 1

See below.

Explanation:

Part A: The point or points of intersection are the values of the variables which satisfy both equations at a particular point.

If `y=2^x` and `y= 4^(x-2)` it follows that `2^x=4^(x-2)` and the solution of which will satisfy both equations. This idea of equating the two equations is something that you would have come across before, maybe without realising it. Consider this:

`x^2+6x+9=0`

This can also be looked at as being the two equations:

`y=x^2+6x+9` and `y=0`

So the points of intersection are the roots in this case and satisfy the equation:

`x^2+6x+9=0`

This explains why it will be the solution to:

`2^x=4^(x-2)` ,because this is the equation formed at the intersection.

Part B:

You just put integer values between -4 and 4 in each equation and when the output is the same for both that is the solution.

`2^(-4)= 0.0625` ,and `4^(-6)=0.000244`

`2^(-3)= 0.125` , and `4^(-5)=0.000977`

`2^(-2)= 0.25` , and `4^(-4)=0.003906`

`2^(-1)= 0.5` , and `4^(-3)=0.01563`

`2^(0)= 1` ,and `4^(-2)=0.0625`

`2^(1)= 2` ,and `4^(-1)=0.25`

`2^(2)= 4` , and `4^(0)=1`

`2^(3)= 8` , and `4^(1)=4`

`2^(4)= 16` , and `4^(2)=16` (This is the solution when x = 4 )

Part C:

To solve graphically you just plot the two equations `y=2^x` and `y= 4^(x-2)` together.

Graph of `y=2^x` and `y= 4^(x-2)`.

You can see the point of intersection is at x =4 and the value of y is 16

Answer 2

Just for interest sake lets derive the actual value of `x` by algebra

Explanation:

There are only two variables

We have `x` which is the independent variable and as such may take on any value we so choose.

The other value of `y` is called the dependant variable which is a fancy way of stating that its is the answer and its value is directly linked to that of `x`

We are given:

`y=2^x" ".....................Equation(1)`
`y=4^(x-2)" "................Equation(2)`

Iff (means if and only if) the two plots intercept at at least 1 point then we may equate them to each other through `y` and it be true. That is, it has a solution.

Write as:

`2^x=4^(x-2)`

Taking logs of both sides (type of log does not matter). I choose log to base e

`ln(2^x)=ln(4^(x-2))`

This is the same as

`xln(2)=(x-2)ln(4)`

`xln(2)-xln(4)=-2ln(4)`

`x=(-2ln(4))/(ln(2)-ln(4))` Multiply RHS by `(-1)/(-1)`

`x=(+2ln(4))/(ln(4)-ln(2)) =4`
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