Asked by jonathan
                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
            
        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
Answers
                    Answered by
            Anonymous
            
    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
    
xln2+xln4=xln2+ln2-xln4-ln4
xln4=ln2-xln4-ln4
(x)2ln4=ln2-ln4
solve for x
                    Answered by
            jonathan
            
    that didn't simplify anything... it says that x=ln(1/5)/ln2
    
                    Answered by
            Reiny
            
    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
    
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
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