Can you Solve Pure Mathematics from Oxford University ? | 8^k+2^k=5

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[@RealQinnMalloryu4]

(K ➖ 3k+2)

[@akeebplazy1124]

Here is a shorter solution, which is more computationally intensive:
8^k + 2^k = 5
(2^k)^3 + 2^k = 5
Let 2^x = y
y^3 + y = 5
y^3 + y - 5 = 0
Using a few test value we can see that the root lies in (1,2)
Now use Newton-Raphson Method with y0 = 1.5
After only a few iterations we get y is approximately = 1.5151598
We can now write y^3 + y - 5 = 0 as
(y - 1.5151598)(y^2 + ay + b) = 0
We can use synthetic division to find out that a is approximately = 1.51598 and b is approximately = 3.2982
So now
(y - 1.5151598)(y^2 + 1.51598y + 3.2982) = 0
Now trying to figure out roots of y^2 + 1.51598y + 3.2982 = 0
We get Discriminant = (1.51598)^2 - 4(3.2982)
Clearly(or with computation), (1.51598)^2 < 4(3.2982)
=>Discriminant is negative
=>There are no real roots of y^2 + 1.51598y + 3.2982 = 0
=>Only real root of (y - 1.5151598)(y^2 + 1.51598y + 3.2982) = 0 is 1.5151598
=>Only real root of y^3 + y - 5 = 0 is 1.5151598
=> 2^k = 1.5151598
=> k = [log(1.5151598)]/log2
=> k is approximately = 0.600255
Which is same as the answer in the video.
This method requires more computation but since calculators are allowed this method is easier. If graphing calculators were allowed then we could easily see that the only real root y^3 + y - 5 = 0 is 1.5151598 . We would have to input y = x^3 + x -5 into the graphing calculator and see when it crosses the x-axis, then we can easily figures out that k = [log(1.5151598)]/log2

[@superacademy247]

I appreciate you sharing your alternative approach! 🔥🔥Thanks for a different perspective! 🔥🔥🔥