X,2x+5=8

Is this a joke? x=1

Too easy

2^25 - 1 = 32Mi - 1

5 sec ready

Alternatively, for the last problem, 500^2 - 499^2 = (500-499)(500+499) = 1(999) = 999

Nice - straight forward without a lot of gibberish

2^x = 3^x is equivalent to x.ln(2) = x.ln(3), so it is equivalent to x = 0. Nothing else!

x = 0.

Without calculation 😂😂😂😂 x=0 😂😂😂😂😂😂😂😂😂😂😂😂😂

2^19(2^1-1)=2^4x=>4x=19 x=19/4

ouch 3^x . 5^x = 15

It's not the original problem: 3^x . 5^x = 145

x=log50/log5=(log5+log10)/log5=1+1/log5 x≈1+1/0.699≈1699/699≈2.43 5^2.43≈49.95

sehr gut :)

5ˣ = 50 ln(5ˣ) = ln(50) xln(5) = ln(50) x = ln(50)/ln(5) ------------------------------------------ And now for a far too detailed solution to the second problem: 2²⁵-1 = (2²⁵ᐟ²)²-1² = (2²⁵ᐟ²+1)(2²⁵ᐟ²-1) = (2¹²⁺⁽¹ᐟ²⁾+1)(2¹²⁺⁽¹ᐟ²⁾-1) = (2¹²*2¹ᐟ²+1)(2¹²*2¹ᐟ²-1) = (4096*2¹ᐟ²+1)(4096*2¹ᐟ²-1) = (4096*2¹ᐟ²)²-1² = (4096*2¹ᐟ²)²-1 = 4096²*(2¹ᐟ²)²-1 = 4096²*2⁽¹ᐟ²⁾²-1 = 4096²*2²ᐟ²-1 = 4096²*2²ᐟ²-1 = 4096²*2¹-1 = 4096²*2-1 = 4096*4096*2-1 = (4000+90+6)*(4000+90+6)*2-1 = [(4000*4000)+(4000*90)+(4000*6)+(90*4000)+(90*90)+(90*6)+(6*4000)+(6*90)+(6*6)]*2-1 = (16000000+360000+24000+360000+8100+540+24000+540+36)*2-1 = (16360000+24000+360000+8100+540+24000+540+36)*2-1 = (16384000+360000+8100+540+24000+540+36)*2-1 = (16744000+8100+540+24000+540+36)*2-1 = (16752100+540+24000+540+36)*2-1 = (16752640+24000+540+36)*2-1 = (16776640+540+36)*2-1 = (16777180+36)*2-1 = 16777216*2-1 = 16777216+16777216-1 = 10000000+10000000+6000000+6000000+700000+700000+70000+70000+7000+7000+200+200+10+10+6+6-1 = 20000000+12000000+1400000+140000+14000+400+20+12-1 = 32000000+1400000+140000+14000+400+20+12-1 = 33400000+140000+14000+400+20+12-1 = 33540000+14000+400+20+12-1 = 33554000+400+20+12-1 = 33554400+20+12-1 = 33554420+12-1 = 33554432-1 = 33554431

Thanks. Also: 5^(x) =[ 5^(2)] * 2, so 5^(x - 2) = 2, so x - 2 = (log2)/(log5), etc.

Why not just : 1 , -1, 0?

I looked at it and thought this makes 0 sense with integers must be 0 idk how harvard students would fail it

5^5 = 5×5×5×5×5 = 3125 , 4^7 = 4×4×4×4×4×4×4 = 16384 = 3125 × 16384 = 51200000 is correct Answer

(x^(1/3))^2 = 16 x^(1/3) = +/- 4 x^(1/3) = 4 yields x = 64 x^(1/3) = -4 yields x = -64

Shut up with your annoying lip smack asmr shit

yeah.. the first problem is completely made up

how is this a challenging problem? and do you have any evidence harvard failed to solve it?

Instead of beating around the bush, just take the log of both sides, and solve the quadratic equation you get. That's it...

Nice

.25

25%

2 to a power is always 4 to half that power. This suggests we try x=4. Sure enough, 4^4 = 2^8 = 256.

I finished school two years ago, but the algorithm suggested this to me and I couldn't solve it in my head so I got curious

As an alternate approach for your first problem. x^2 - y^2 = 28 xy = 48 12x^2 - 12y^2 = 336 7xy = 336 12x^2 - 12y^2 = 7xy 12x^2 - 7xy - 12y^2 = 0 12x^2 - 16xy + 9xy - 12y^2 = 0 4x(3x - 4y) + 3y(3x - 4y) = 0 (4x + 3y)(3x - 4y) = 0 Case 1: 4x + 3y = 0. So 4x^2 + 3xy = 0, and therefore 4x^2 + 3(48) = 0. The minimum possible value for the LHS is 144, which is bigger than 0, so no real solutions. Case 2: 3x - 4y = 0. Multiplying through by x, we get 3x^2 - 4xy = 0. Since xy = 48, this means that 3x^2 - 4(48) = 0, or x^2 = 64, so x = +/- 8. y is then +/- 6. These do work with the original equations, so we have solutions.

X= 8; y= 6

X=6 y=8

Why lie? 4^x = (2^2)^x = 2^(2x) = 8 = 2^3, 2x = 3, x = 3/2.

0

Who is einstein? your dog?

What nonsense, this is an extremely simple logarithmic equation. FWIW, I had an algebra teacher who used to make us chant, "A logarithm is an exponent." Once you understand that, just convert those words to mathematical terms. x is the exponent, "is" means =. And since the base of an exponent is the thing "holding the exponent up" if you will, then 4 is the base. So you have x = log base 4 of 8. Not that that helps to solve it by hand, but you can now look up the solution in a log table. I didn't even watch this video, but it's obvious you use exponent rules to rewrite this as 2 squared to the x = 2 cubed. Since a power raised to a power is the product of the two powers, it's a snap to write 2 to the 2(x) = 2 to the third. So obviously, 2x = 3, and any 6th grader knows how to solve THAT.

Einstein Failed to Solve This - LOL

FITH GRADE stuff! LOL

huh? 2^2x=2^3, 2x=3, x=1.5

2^3 = 8, 2 = 4^.5 so substitute 4^.5 for 2 and you get 4^(.5)3 or 4^1.5 = 8. Or Rewrite the equation as 2^2x = 2^3 and solve for x

Too complicated

Seriously? This title is so idiotic that I'm going to block your channel.

Reported for misleading title.

Hi there, in the -b formula, -b should be substituted with positive six as the b is negative 6. Regards

Einstein did not fail to solve this, you really think the mind behind general relativity couldn't solve a basic logarithm problem?

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0.25

ans 1

Yeah no.