วีดีโอ

Bina Log ke bhi to easily solve kar sakte the na !!! (4^b) = 8 [(2^2)^b] = (2^3) => [2^(2*b)] = (2^3) Same base hone ke karan, Powers ki Value bhi same hongi. Therefore, (2*b) = 3 b = 1.5 That's easier ! And check ke lie : 9^2 - 16^1.5 = 17 81 - [(4^2)^1.5] = 17 81- (4^3) = 17 81-64 = 17 17=17

5,3 / 3,5

√x=e^{iπ} and x=[e^{iπ}]^2=e^{i2π}=1. √x=√1=±1, so select the negative root to agree with the given equation.

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nice solution

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This can literally be solved in 2 lines. 5 ^(2a-2) = log[2,5] { read as log 2 to the base 5 } => a = 1+0.5 log[2,5] done.

1/7^1/x=7 (7^-1)^1/x=7 7^-1/x=7^1 -1/x=1 x=-1/1 x=-1

Good explanation ❤❤

4^x=12 Taking both side log Log4^x=log12 xlog4=log12 Both side divided by log4 xlog4/log4=log12/log4 X=log12 on the base 4 To check 4^x=12 4^log12 on the base 4=12 (a^log b on the base a= b) So, 12=12

😊sime way that take sqroot of both sides then solve linear

√(x+4)2=√4*4 X+4=4 X=0

Great

4^×=4(20)=>4^×-2=5 =>(×-2)(log 4)=log5 =>×=2+log@4 (5)

बड़े अक्षरों में लिखने का प्रयत्न करो।।😂😂😂

x = log3 to base 15 ❤

x = 1 + ln3/ln15

Thanks

3^0=5^0=6^0=7^0 x=0

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thanks🌹

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Inta karne ki kya jaroorat... Kisi bhi number ki power zero hai to value 1 aata hai

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super🤗🤗🤗

If m and n are supposed to be positive integers, then the unit digit of 7^m could be one amongst {1,3,7,9} and the unit digit of 5^m is always 5. The unit digit of the number 2376 is 6. Getting 6 as unit digit will only be possible if unit digit of 7^m is 1 (it will be 11-5=6). Hence m must be a multiple of 4 to have unit digit of 7^m as 1. (e.g. 7^4 = 2401, 7^8 = 5,764,801, etc.) Assume m = 4k. Then 7^m = 7^(4k) = (2401)^k. Now, if we take k=1 and n=2, then we get 2401-25=2376. Thus, m = 4k = 4 x 1 = 4 and n= 2 is the solution.

From begin you make 1 error, you don't require that 6-x must be grater or equal to zero, because in the other side sqrt(x) in equation will be negative value.

Look for smallest integer power of 7 greater than 2376: m=ceiling(ln2376/ln7)=4. 7^4-2376=25=5^2=5^n. n=2.

Nonsense 3/x . 3/x = x/3 3.3.3 = x.x.x 3=x

X is 3

3/x X 3/x = x/3 3 X 3 X 3 = x X x X x 3 = x

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3^{x+2}+3^x=3*18=54; (9+1)*3^x=10*3^x=54; 3^x=54/10=5.4; x=ln(5.4)/ln3=1.53503...

x=ln12/(3ln3)=0.75395...

Mashallah sir ji

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Nice 👍👍

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-1/3

25ⁿ - n⁴ = 609 (5²)ⁿ - n⁴ = 625 - 16 {(5ⁿ)² - (5²)²} - {(n²)² - (2²)²} = 0 (5ⁿ + 5²)(5ⁿ - 5²) - (n² + 2²)(n² - 2²) = 0 (5ⁿ + 5²)(5ⁿ - 5²) - (n² + 2²)(n + 2)(n - 2) = 0 (5ⁿ + 5²)(5ⁿ - 5²) = (n² + 2²)(n + 2)(n - 2) F(n) = (5ⁿ + 5²)(5ⁿ - 5²) G(n) = (n² + 2²)(n + 2)(n - 2) n² + 2² = 0 n² = - 2² n = ± √(- 2²) = ± √(- 1 * 2²) = ± √(i² * 2²) = ±2i <--- not real solution F(2) = ( 25 + 25)( 25 - 25) = 50 * 0 = 0 G(2) = (2² + 2²)(2 + 2)(2 - 2) = 0 F(-2) = (1/5² + 5²)(1/5² - 5²) = (1/5² + 5²*5²/5²)(1/5² - 5²*5²/5²) = (1 + 5⁴)(1 - 5⁴)/5² < 0 G(-2) = ((-2)² + 2²)((-2) + 2)((-2) - 2) = 0 F(2) = G(2) ∴ n = 2

❤❤❤❤ mashallah ♥️❤❤

If you are going to assume x is a positive integer, then x is less than or equal to 2. x = 2 is a solution.

Thanks ❤🎉

Mashallah sir ji ❤❤❤❤❤

X1 =0, X2=[3+(5)^(1/2)]/2, X3=[3-(5)^(1/2)]/2.

i=e^{iπ/2}; √i=±√(e^{iπ/2})=±e^{iπ/4}=±[cos(π/4)+i*sin(π/4)]=±(√2/2+i√2/2)