I've been working on this problem all night and I can't seem to do it. Can someone give me a hint on how to start it?

4^x - 10 * 4^-x = 3

Thanks.

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- October 29th 2007, 10:37 PMktcrewExponential Equation
I've been working on this problem all night and I can't seem to do it. Can someone give me a hint on how to start it?

4^x - 10 * 4^-x = 3

Thanks. - October 29th 2007, 11:54 PMticbol
Let me do it all the way to avoid too much asking/answering if in case.

**4^x - 10 * 4^-x = 3**

If that is

(4^x) -10(4^(-x)) = 3,

then,

4^x -10/(4^x) = 3

Clear the fraction, multiply both sides by 4^x,

4^(x+x) -10 = 3(4^x)

4^(2x) -3(4^x) -10 = 0

That is a quadratic equation in 4^x,

[if y = 4^x, then that is y^2 -3y -10 = 0]

so, factoring it,

(4^x -5)(4^x +2) = 0

When 4^x -5 = 0,

4^x = 5

Take the common logs of both sides,

xLog(4) = Log(5)

x = log(5) / Log(4)

x = 1.160964047 ---------------answer.

When 4^x +2 = 0,

4^x = -2

Take the common logs of both sides,

xLog(4) = Log(-2)

There are no real number logs of negative numbers, so reject 4^x +2 = 0. - October 30th 2007, 05:30 AMktcrew
Thankyou. (Handshake)

So we always get rid of a negative first? Or was that just the easier way of doing that problem? - October 30th 2007, 10:14 AMticbol
Get rid of the fraction as soon as possible is more like it.

The "negative", or negative exponent, means it should be in the denominator in that case.

Easier way of doing that problem? I don't know.

"Easier" is relative. It might be easier for you but it might not be for others.

In my case, I just solve any Propblem in any way I can. Easy or not easy way. For me, the solution is more impportant than the elegance of the solution.