Solve

4^x + 128 = 9.2^x+2

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- August 26th 2009, 10:27 PMdeej813how do you solve for x when x is a power?
Solve

4^x + 128 = 9.2^x+2 - August 26th 2009, 10:42 PMnikhil????
is it (9.2)^x or 9(2)^x

- August 26th 2009, 10:45 PMdeej813
its (9.2)^x+2

- August 26th 2009, 11:06 PMkhotso
here is my try

4^x + 128 = 9.2^x+2

arrenge the eguatio to get

4^x-9.2^x=-126

9.2^x+4^x=126

take logarithm

log(9.2^x)+log(4^x)=log126

xlog9.2+xlog4=196

x(log9.2+log4)=196

x=196/(log9.2+log4) - August 26th 2009, 11:08 PMdeej813
sorry its (9.2)^(x+2) x+2 is the power

could you possibly do it without logs please because i havn't learnt them yet - August 27th 2009, 04:08 AMaidan
- August 27th 2009, 05:46 AMpacman
**deej813**, is this the equation that you want to be solved?

http://www4a.wolframalpha.com/Calcul...image/gif&s=38

i don't know how to deal with this without resorting to graphing.

http://www4a.wolframalpha.com/Calcul...image/gif&s=38 the blue line is the graph of the left hand side (http://www4a.wolframalpha.com/Calcul...image/gif&s=56)

the red line is the graph of the right hand side (http://www4a.wolframalpha.com/Calcul...image/gif&s=12)

the intersection of the two lines is the solution, x = 0.191

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But what if what HE meant was http://www4a.wolframalpha.com/Calcul...image/gif&s=23 ?

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Integer solutions are x1 = 2 and x2 = 5.

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(Shake)(Clapping)(Clapping)(Shake) (Wait)

http://www4a.wolframalpha.com/Calcul...image/gif&s=23

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Again, the blue thread is the left hand side of the equation,

the red thread is the right hand side of the equation, the intersection of the two is the solution. Will they ever meet?