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Bro from the equation we can get x=log(base5)40 which is x=1+log8

2^103=10141204801825835211973625643008 (10 nonillion, 141 octillion, 204 septillion, 801 sextillion, 825 quintillion,835 quadrillion,211 trillion, 973 billion, 625 million, 643 thousand and 8! Could not resist. 😁

surely just do x=log base 4 of 28?

Very nice explanation

I watched it at 2am out of a random recommendation and thoroughly enjoyed it

i watched it drunk out of a random recommendation and thoroughly enjoyed it

{2^a}^3 = 8^a = 4^a + 2^a = {2^a}^2 +{2^a} {..}=: x <> 0 -> x^2 = x + 1 2^a = x_0 -> a = lb(x_0)

you mean 2^97

Wouldn't it be easier to divide by 2ª instead of 8ª ?

x = ln(18)/ln(9) = ln(2*9)/ln(9) = (ln(2) +ln(9)) / ln(9) = 1+ln(2)/ln(9)

Thanks for the explanation 🌹

This is incorrect

i have put x =logytobase5 to solve this. it gave 5 power ((root5 + 1 )/2)

X = log(44) / log(11)

its 1

2^96 + 1 = 2^97 right? am i dumb?

What is a logarithm?

तुम्ही x ची किंमत सांगितलीच नाही.

हे चूक आहे.

making very complicated

2⁹⁶+¹ =2⁹⁷ in 3:36

Umm... 9×3=27

Bro it was incorrect

Kya bawaseer hain??? Galat kar diye beech mein. It was log3/log6.

pls dont teach us incorrect math problems

Please don’t do incorrect math solutions

Ye kya tha bhai

36is not equal to 9×3 wtf

wow great from uzbekistan

1 is close enough for practical purposes

Thanks for getting to it rather than a lot of useless talk like the other guy.

Mike is not good

Please work on your math skills, 1st one is fully wrong.

X = 1. That makes it 8 + 2 = 10. 2 to the exponent 3(1) is 8.

Can't you Just use last fermat teorem and check only for n=2?

cool video

Thanks a lot

Why don't you just use log from first

that's not "X" That's "n"

Or you could divide by 2^a at the beginning, avoiding the fractions. But it is a neat method, to turn it into solving a quadratic.

Thanks a lot

2 1/2?

On peut arriver directement au meme resultat sans diviser par 16/x

7^x = 28 Take log_7 of both sides. x.= log_7(7 * 4) x = 1 + log_7(4) or x = log_7(28) These log questions are absurdly easy.

Answere is incorrect

t=7^x t^2-t-1=0 t=7^x=(1+rt5)/2 x=log_7[(1+rt5)/2] t>0

Why all the fuss to get started? Just recognize the given equation as equivalent to (2ˣ)⁰ + (2ˣ)¹ = (2ˣ)² which rearranges as (2ˣ)² - (2ˣ)¹ - (2ˣ)⁰ = 0 and is solved as (2ˣ) = ½ ⋅ ( 1 ± √5 ).

b^2-4ac=D

Impressive the strength you use to write, I'm also this way hahaha

=2.2^48=1×2^49ans