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3 log10 ^=600. Log10^. =200. Xlog10. =Log200. X=log200/log10. X=log2+log100=log2+2

Nice! Thanks!

Why not going straight to the point? 10^x+10^x+10^x=600 3*10^x=3*200 10^x=200 10^x=2*10^2 10^(x-2)=2 x-2=log10(2) x=log10(2)+2

Well explained

2 x 5^k = 100 5^k = 50 5^(k-1) = 10 (K-1) log5 = log 10 = 1 K = 1 + 1/log5

Perfect method to learn and understand

Good morning maam

❤❤❤❤❤

Different but nice solution

❤very good solution

nice explaination

Very well explained

Great 😊

Perfect teaching

Good teaching skills

K=2.431?

x + y = 20 x*y = 44 y = 20 - x x*(20-x) = 44 20*x - x^2 = 44 x^2 - 20*x + 44 = 0 Note that this quadratic contains essentially the given problem - the roots of the quadratic are the numbers we seek. Because of symmetry, it doesn't matter which is which. The result is x = 10 + 2*sqrt(14), y = 10 - 2*sqrt(14) or vice versa.

me: x=10 y=5

how you decide to re right (1series) terms in 99999/9, (i mean there are different ways like 8888/8). So my question is why terms of 9 not other?

Nice algebra questions solved ❤

Why does no one use this simple substitution method? Substitute x=10+z and y=10-z into the second equation and rearrange to 100-z^2=44 or z^2=56, which has roots z=±2√14. Thus x=10±2√14 and y=10∓2√14.

Nice explanation

Very well

А почему сразу не логарифмировать по базе 5?

y=150/x x+150/x=15 x²+150=15x x²-15x+150=0 x²-15x+225/4+375/4=0 (x-15/2)²=-375/4 x-15/2=+-5√15i/2 case 1 ; x=5√15i/2+15/2 , y=-5√15i/2+15/2 || case 2 ; x=-5√15i/2+15/2 , y=5√15i/2+15/2

i beat 56 secs

The General and easiest approach 👍🖐️♥️

Nice teaching skills

Very well taught maam

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Great😂

very nice

Good

2/-2

BLO IT'S ±2!!!! WHY YOU GOING SO COMPLEX???!

Thats not an 8 Minute problem.

I have a feeling about this one😊😊

Nice way to explain

Very well explained

Perfect olympiad question here, thanks

I respect the one who is giving us the knowledge in the video🖖🖖

This is so good

Oh my, the explanation was simply exquisite 💐

Been looking for this since 2 and half hours😮😮

love it

2(5^k) = 100 5^k = 50 k (log5) = log50 k = log50/log5 = 2.4307

Wrong answer madam! There are 6 set of solutions for(a, b, c)

No solution

I understend nothing, where is proof? imho it's zero b/c 9^4 will never be div by 2 and be natural so only solution zero, or there is no solution b/c 9=/=4

2x⁴=32 x⁴=16=2⁴ So x=2