find the points of intersection of the following algebraically
y=2^x + 4^x
y=2^x+1 - 4^x+1
I set them equal to each other but not sure if you use logs or what to solve it
xln2+xln4=(x+1)ln2-(x+1)ln4
xln2+xln4=xln2+ln2-xln4-ln4
xln4=ln2-xln4-ln4
(x)2ln4=ln2-ln4
solve for x
that didn't simplify anything... it says that x=ln(1/5)/ln2
anonymous is wrong
you have to take ln of both sides, not the individual terms.
2^x + 4^x = 2^(x+1) - 4^(x+1)
2^x - 2^(x+1) = -4^x - 4^(x+1)
2^x(1 - 2) = -4^x(1+4)
2^x = (5)4^x
2^x = 5(2^(2x))
divide both sides by 2^x
1 = 5(2^x)
1/5 = 2^x
now ln both sides
ln(1/5) = xln2
x = ln(1/5)/ln2
To find the points of intersection between the two equations, you need to set them equal to each other and solve for x. Let's start by setting the two equations equal to each other:
2^x + 4^x = 2^x+1 - 4^x+1
To solve this equation, we can simplify it by re-arranging the terms and combining like terms:
2^x + 4^x = 2 * 2^x - 4 * 4^x
Now, we can move all the terms to one side of the equation:
2^x + 4^x - 2 * 2^x + 4 * 4^x = 0
Next, let's factor out a common factor of 2^x:
2^x(1 + 2^x) - 4^x(2 - 4^x) = 0
Now, we have factored out the common factor in each term. To simplify it further, we observe that 2^x is common to both terms, so we can factor it out:
2^x[(1 + 2^x) - 4^x(2 - 4^x)] = 0
Now, we have two cases to consider:
Case 1: 2^x = 0
If 2^x = 0, then x must be a negative infinity since 2 raised to any power will never equal zero. However, negative infinity is not a point of intersection that we can plot on a graph.
Case 2: (1 + 2^x) - 4^x(2 - 4^x) = 0
In this case, we need to solve the equation (1 + 2^x) - 4^x(2 - 4^x) = 0 numerically since it cannot be easily solved algebraically.
You can use numerical methods like graphing the two equations and finding the points where they intersect, or using a graphing calculator or a computer software that can solve equations numerically.
By solving this equation numerically, you will find the values of x that correspond to the points of intersection between the two equations.
Remember that finding numerical solutions may not always be as straightforward as solving algebraically, but using numerical methods can help you find accurate solutions.