With a Little Help from MathQueen!

First, the actual video:

Can you crack this beautiful equation? – University exam question
Math Queen | 7 Dec. 2024
https://www.youtube.com/watch?v=ecYy-3gxzhU

Then, my simplified resumé of the steps:

2^x + 8^x = 130
2^x + (2³)^x = 130
2^x + (2^x)³ = 130
2^x = y
y + y³ = 130

5 = y?

5 + 125 = 130!

5 = y, yes!

2^x = 5.
log(2^x) = log(5)
x*log(2) = log(5)
x = log(5)/log(2)

And now, some application to Carbon dating. How do I exactly derive a time implication from a pmC value or perhaps better its decimal counterpart. Imagine I want to know how much 70 pmC is worth, or 0.7 times the "modern carbon [14] level corrected for pre-industrial values"? Can I do this in this way?

0.5^(x/5730) = 0.7
log(0.5^(x/5730)) = log(0.7)
x/5730 * log(0.5) = log(0.7)

x = 5730 * log(0.7) / log(0.5)
x = 2948.5042803145152554

0.5^(2948/5730) = 0.7000427025825161
0.5^(2949/5730) = 0.699958024872583

YES!

Thank you very much!/HGL

PS, let's test it for a calibration. I want to know, what year and what pmC I need to get 4000 BC or as close as possible. I'll ignore the "graph curvature" of the surrounding plots. It's on the III—IV table, so, revised:

2039 BC

78.209 pmC, so dated 4089 BC

2022 BC

79.035 pmC, so dated 3972 BC

It looks like I need two or three of the latter to balance the former.

(2039 + 2022 + 2022) / 3 =
(78.209 + 79.035 + 79.035) / 3 =

(2039 + 2022 + 2022 + 2022) / 4 =
(78.209 + 79.035 + 79.035 + 79.035) / 4 =

Now, let's calculate:

(2039 + 2022 + 2022) / 3 = 2028 BC
(78.209 + 79.035 + 79.035) / 3 = 78.76 pmC or 0.7876

x = 5730 * log(0.7876) / log(0.5)
x = 1974 (extra years)

1974 + 2028 = 4002 BC

2039 BC

78.209 pmC, so dated 4089 BC

2028 BC

78.76 pmC, so dated as 4002 BC

2022 BC

79.035 pmC, so dated 3972 BC

Thank you again, Susanne!/HGL

PPS, I note that fractions are necessary, replace them with pmC values in the percent form, and it won't work./HGL