Two questions about indices?

Question

Hi!

I´ve been working on indices today and two questions have arisen.

1. For the following kind of problem for example

x^(1/2) - 5x^(1/4) + 6 = 0

I know how to solve this problem, by letting y = x^(1/4) and then factorising. But how come when I try to solve these kind of problems by writing out the problem as for example √x - 5∜x + 6 = 0 the answer is never correct? Am I doing something wrong or is it not possible to work it out this way?

2. Finally, is there a method to solving the following kind of problem or is it only through observing and trial and error? If there is a method please explain :)

8^x = 4^(1-x)

Thank you in advance for your time and help! Cheers!

Answer 1

2. Finally, is there a method to solving the following kind of problem or is it only through observing and trial and error? If there is a method please explain :)



8^x = 4^(1-x)



Both "8" and "4" are BASE-2 numbers...



8 = 2^3

4 = 2^2



so, replace and all will become CLEAR !!



(2^3)^x = (2^2)^(1 - x)

2^(3x) = 2^(2 - 2x)



if the bases are equal, then the exponents must be EQUAL...

3x = 2 - 2x

5x = 2

x = 2/5

check...I did



one at a time

Answer 2

1. You probably did something wrong; it shouldn't matter if you substitute y = ∜x, or leave ∜x in the equation.



√x - 5∜x + 6 = 0

√x - 3∜x - 2∜x + 6 = 0

∜x(∜x - 3) - 2(∜x - 3) = 0

(∜x - 2)(∜x - 3) = 0

(∜x - 2) = 0 or (∜x - 3) = 0



∜x = 2 or ∜x = 3

x = 16 or x = 81





2. It doesn't matter if 8 and 4 are both powers of 2 or not.



8^x = 4^(1-x)



8^x = 4^1 / 4^x



4^x * 8^x = 4



(4*8)^x = 4



x = log 4 / log 32 = (2 * log 2)/(5 * log 2) = 2/5.





Alternative method:

take the logarithm of left-hand and right-hand side.



8^x = 4^(1-x)

x ln 8 = (1-x) ln 4



solve as linear equation in x:



x ln 8 + x ln 4 = ln 4

x = ln 4 / (ln 8 + ln 4) = 2*ln2/(3*ln2 + 2*ln2)

x = 2/5



If the equation had been, say, 13^x = 5^(1-x), then by these same two methods we would have solved the equation and found x = log 5 / log 65.

Answer 3

√x - 5∜x + 6 = 0



Let u = ∜(x)



u² - 5u + 6 = (u - 2)(u - 3) = 0



u = 2, 3



∜(x) = 2 ⇒ x = 2⁴ = 16

∜(x) = 3 ⇒ x = 3⁴ = 81



********************************

8^x = 4^(1 - x)



Use the fact that 8 = 2^3 and 4 = 2^2



2^(3x) = 2^(2 - 2x)



Take base 2 logs of both sides



3x = 2 - 2x



5x = 2



x = 2/5