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.
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.