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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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.
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.
Thankyou. (Handshake)
So we always get rid of a negative first? Or was that just the easier way of doing that problem?
Get rid of the fraction as soon as possible is more like it.Quote:
Originally Posted by ktcrew
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.