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(2²)^x = 14 (2x)(ln 2)= ln 2 + ln 7 x = (ln 2 + ln 7)/(2ln 2) x = 1.9036774610288... Check: 4^1.9036774610288... = 14 Note: (ln 2 + ln 7)/(2ln 2) = (1/2)(1 + log_2 7)

8 degreeded of 5 are 8 x 1,444 x 1,444 x 1,444 x 1,444 x 1,444 = 50,225

x=log200/log8=(log2+log100)/3log2=⅓+2/3log2 x≈⅓+2/(3×0.301)≈767/301≈2.548 8^2.548≈200.02

If you know that the sum of the 2 is 520 you can conclude that x can't be a very high number. Make x equal to 1 that gives 8+2= 10, make x equal to 2 that gives 64+4= 68. Then make x equal to 3 and that gives 512+8= 520. Just use common sense. Your way the chance of making a mistake on the way is very high.

Let x=8^a. Then 1 + x = x^2 or x^2 -x -1 = 0. Thus, x = (1+sqrt(5))/2 and a = ln(x)/ln(8).

No comment 😮

8²+2²=64+4=68 So k=2

Very good 😅

x³=a a³+a²=36 3³+3²=27+9=36 a=3 x³=3 x=cbrt3 ...and, probably, eight complex roots else Which look something like this: cbrt3×cos(n×2pi/9)+i(cbrt3)×sin(n×2pi/9) n= 1, 2, ... , 8

x=(log(sqrt5+1)-log2)/log3 x≈0.438 9^0.438-3^0.438≈0.99993

x=(log(sqrt5+1)-log2)/2log2=log(sqrt5+1)/2log2-½ x≈0.347 1+4^0.347≈2.618 16^0.347≈2.617

a=log(sqrt5+1)/log2-1 a≈0.694

5^2×=28 5^2×=2×7×2 5^2× = 4×7 2xlog5=log4+log7 2x =log4/log5 + log7/log5 2x=log4(5)+log7(5) x=1/2(log4(5)+log7(5)

Make it simple, 1 to the power of anything is just 1.

Nice thanks 😊

There are easier solutions.

x=(log(sqrt5+1)-log2)/2log2=log(sqrt5+1)/2log2-½ x≈0.347 1+4^0.347-16^0.347≈0.0006

2^x=a a³+a²=36 36=2²×3² 3³+3²=27+9=36 a=3 2^x=3 x=log3/log2=lb3 x≈0.477/0.301≈1.585 8^1.585+4^1.585≈36.0026

k=log(sqrt5+1)/log2-1 k≈0.694

x=log35/log7=(log7+log5)/log7=1+log5/log7 x≈1+0.699/0.845≈1544/845≈1.827 Deviation is about 0.0001 7^(1.827)≈34.994

Its three not thiri

i used the calculator

x=(2log2+log7)/2log5 x≈(2×0.301+0.845)/(2×0.699)≈1447/1398≈1.0351 Deviation is about 0.0002 5^(2×1.0351)≈27.99

x=(3log2+log3)/2log5 x≈(3×0.301+0.477)/(2×0.699)≈230/233≈0.9871245 Deviation is about 0.0002 5^(2×0.9871245)≈23.99

(5^x)² = 10 /√ 5^x = √10 What are you yappin about for 8 minutes💀

x=log10/2log5=1/2log5 x≈1/(2×0.699)≈500/699≈0.71531 Deviation is about 0.00003 5^(2×0.71531)≈9.9991

x=(log(sqrt5+1)-log2)/log2=log(sqrt5+1)/log2-1 x≈0.694242 1+2^0.694242-4^0.694242≈-0.0000002

easy you are doing to too much toroghly do it small

a=(log(sqrt5+1)-log2)/log3

x=(log(sqrt5+1)/log2)-1 x≈0.6942

5^x=a a+a²=a³ 1+a=a² a²-a-1=0 a=(sqrt5+1)/2 5^x=(sqrt5+1)/2 x=(log(sqrt5+1)-log2)/log5 x≈0.299 5^0.299+25^0.299-125^0.299≈-0.00006

X^2(x+1)=9^2x10 ×+1=10 / ×^2=9^2 ×=9

why didnt we cancel (x-3) 5:10

X=9

Its very easy when you realize 1^x is just 1, and 100^x = 10^(2x)

x*x*(x+1)=810 9*9*10=810 x=9

2^x=a a²-a-1=0 a=(sqrt5+1)/2 2^x=(sqrt5+1)/2 x=(log(sqrt5+1)-log2)/log2=(log(sqrt5+1)/log2)-1 x≈0.69424. 4^0.69424-2^0.69424-1≈-0.000005.

🔥

8^k+2^k=30 2^k=x x³+x-30=0 With Horner's scheme find root x=3 2^k=3 k=log3/log2 k≈0.477/0.301≈1.585.

x=log15/log2=(log3+log5)/log2≈(0.477+0.699)/0.301≈1.176/0.301≈3.907. Deviation is about 0,0001.

2^×=15.... =>×=log@2 15

2ˣ=15 log₂(2ˣ)=log₂15 x∙log₂2=log₂15 x=log₂15

This also works, of course x=Log(2) 15; x= 3.9069

Thx for making it far more difficult than the og problem.... go study some more

Get a better pen next time... 4/10 FAIL

(5^x)+(10^x)=20^x (5^x)[{(2²)^x}-(2^x)-1]=0 (2^x)²-(2^x)-1=0 a quadratic equatiom in 2^x 2^x=½[1±sqrt(5)] 2^x=½[1+sqrt(5)], a golden ratio =ß say --> x=[log(ß)]/log(2) or 2^x=½[1-sqrt(5)], conjugate ß = -(1/ß) rejected as 2^x can't be negative

X= {log (*5-1)/2}/{3 log (1/2)}......May be Explain later

Ridiculous

Dare to solve : x^5 - x^(5 - x) = { (x + 1)x^(x + 1) }^x-1

X= log₂((1+√5)/2)