Re: [Math4u] equation

Say f(x) =5^x-4^x-3^x-2^x.-1^x

One method is you could graph f(x) and draw a line on the graph for f(x) = 25 and read off the point.

Another method is you do a Newton's approximation, my favorite.

One guess is that you can see that you have 5^x on one side and 5^2 on the other side so guess 2. But then notice that all the other terms are negative so this tells us there is only one solution.

another method is to rearrange the formula with the most important x on the left.

5^x = 5^2 + 4^x + 3^x + 2^x 1^x

let x=0, then

5^x = 5^2 + 4^0 + 3^0 + 2^0 +1^0 = 25+1+1+1 +1 =29 so x=log30/log5=2.11328275255938

let x=2, then

5^x = 5^2 + 4^2 + 3^2 + 2^2 +1= 25+16+9+4+1=55 so x=log55/log5=2.48989610240498

What we want is for the x we put in on the right hand side of the equation to equal the x we calculate on the left-hand side of the equation. In this case, if we keep putting the calculated value on x back in on the right side of the equation, we will quickly get to a final answer that does not change.

Alternately, we can use Newton's approximation to choose the next guess. We take the input x minus the output x for the last two approximations. We calculate the change in this discrepancy verses change in assumed x. This gives us a slope. We use this slope to calculate the next input x. This will get us the answer very quickly.

Regards,

Brian Jensen

 

 

----- Original Message -----

From: funkclear

To: Math4u@yahoogroups.com

Sent: Monday, October 08, 2007 11:12 AM

Subject: [Math4u] equation

How would you solve this equation?
5^x - 4^x - 3^x - 2^x - 1^x = 5^2

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