College algebra recapitulation help?

Question

Earlier today there were three questions by one asker:
1. Given a function y=f(x) and a point on it (4,5), then what is the corresponding point on y=2f(x-6)
2. If f(x) = 1/x^2 - 2x + 8, find where f(x) < 0; this one I could do, but I ran into a cubic equation and have forgotten how to solve/factor those
3. Solve for x: 4^(2x) + 1 = 2^(3x) + 6
On this last one, the one that has driven me most crazy, I was using log to the base 2 and some such; there has to be a more elegant way to solve these.

Answer 1

1. vertical stretch by a factor of 2, and horizontal translation to the right by 6 units

(4, 5*2) = (4, 10) after stretch

(4 +6,10)= (10,10) after translation final answer

3. LN= Natural log log to base episolon, can find in most calculators

2xLN(4) + LN(1)= 3xLN(2) + LN(6)

2xLN(4) - 3xLN(2)= LN(6) - LN(1)

x(2LN(4) -3LN(2))=LN(6) - LN(1)

x= LN(6) - LN(1)/(2LN(4) -3LN(2))

Simplification using log laws:

LN(a) - LN(b) = LN(a/b)

As such

x= LN(6/1)/ Ln(16/8)= LN(6)/ LN(2)

2. Use a graphing calculator, takes way long to do sorry

Answer 2

The first one is scaled vertically by 2 and shifted right 6 units, so the point is (10,10)

The 2'nd one is < 0 anywhere that x² - 2x + 8 < 0 or where
(x-4)(x+2) < 0 or -2 < x < 4

Re write the 3'rd one as 2^(4x) + 1 = 2^(3x) +6 so
2^(4x) / 2^(3x) = 5 / 2^(3x) or
2^x = 5*2^(-4x) take the log of both sides to get
x * log(2) = log(5) - 4x * log(2) so
5x = log(5)/log(2) and
x = log(5) / (5 * log(2)) or
x = 0.46438561897747246957406388589788


Doug

Answer 3

2) I don't think that you get a cubic:

mult by x^2
1-2x+8x^2 < 0 (since x^2 is positive, inequality is unchanged)

This is a quadratic!

( I'm assuming that by 1/x^2 you mean 1/(x^2), but it doesn't matter since (1/x)^2 = 1/(x^2)