Simplifying an Exponential Expression

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[GirishManjunathMusic]

Given:
12↑((x – 1)/x) = 3
To find:
8↑x
Separating the fraction:
12↑(x – 1/x) = 3

Using a↑(b – c) = (a↑b)/(a↑c):
12/(12↑1/x) = 3

As 12↑a is never 0, multiplying both sides by 12↑1/x and dividing both sides by 3:
4 = 12↑1/x

As x cannot be 0 by definition, taking both sides to the xth power:
4↑x = 12.

Taking both sides to the 3/2th power:
(4↑x)↑3/2 = 12↑3/2

Using (a↑b)↑c = a↑bc = a↑cb = (a↑c)↑b:
(4↑3/2)↑x = 12↑3/2

As 4↑3/2 = 8:
8↑x = 12↑3/2.

EDIT: 12↑3/2 == (2√3)³ = 24√3.

[spelunkerd]

Wow, what a great exercise to practice exponential gymnastics. I used third method, with a few long cuts, ha ha.

[ManjulaMathew-wbzn]

I like 3rd method the best.
First method had an opportunity to make simpler.
x=log base4(12)
By definition 4^x=12. And the rest is the same as other 2 methods. Basically the variable a you used is the same as x.

[iabervon]

I used the second method, but I handled 4^x=12 a little differently. If 4^x=12, 2^(2x)=12. Then 8^x=2^(3x)=12^(3/2).

[_wahahaha]

3rd method is my first method just after seeing this question

[enzogcol]

Really liked it, gonna show to my friends actually!

[rob]

(x-1)ln 12 = x ln 3
x ln 12 - ln 12 = x ln 3
x(ln 12 - ln 3) = ln 12
x ln 4 = ln 12
x = ln 12 / ln 4
= 1 + ln 3 / ln 4

[prateek.]

are you crazy, i literally squeeezed my entire brain for about 2 hours snd still managed to get the answer by only 1 method, and you r solving it by 3 methods. cmon, man you are genius!!

[laurentthais]

Chess mode solving : 4^x = 12, so 2^x = sqrt(12), and 8^x = 12^(3/2)

[hectordpg]

Too fast and difficult for my level but I'll keep trying! 🧐

[mega_mango]

I love how you do not just solve problems, but also do it with different methods of solving so that viewers can see the problem from different sides.

[scottleung]

I used a method similar to your second one.

[jackknopf]

SQRT(4a) = 2a ??

Shouldn’t it be: SQRT(4a) = 2SQRTa ??

[KM-omhm]

I used the second method to solve it, I'm tired of taking logarithm each side always.

[drozfarnyline]

I used both 1st and 3rd partially
Nice one BTW