MIND-BLOWING Algebra Secrets for Easy Problem Solving

[@Quest3669]

Eqn turns to
(1-2x)(2xlog2+log5)= 0. Or
X= 1/2; -1/2 [log 2_ (5)]

[@dan-florinchereches4892]

lg(16^(1-x)*5^(1/x))=lg100
(1-x)*(4-4lg5)+1/x*lg5=2
-x^2*4*(1-lg5)+(4-4lg5-2)x+lg5=0
Since the original equation is 2^(4-4x)*5^(1/x)=2^2*5^2 and 4-4x=2 and 1/x=2 have the same solution 1/2 we can factor it out of the quadratic
x*(2x-1+1)*2(1-lg5)-2x(1-2lg5)-lg5=0
(2x-1+1)(2x-1)(1-lg5)-x(2-4lg5-2+2lg5)-lg5=0
(2x-1)^2(1-lg5)+(1-lg5)(2x-1)+(2x-1+1)lg5-lg5=0
(2x-1)^2(1-lg5)+(2x-1)*1+lg5-lg5=0
(2x-1)((2x-1)(1-lg5)+1)=0
so x2=1/2-1/(2-2lg5)=1/2(1-1/lg2)=-lg5/lg4 where lg is the decimal logarithm

[@roberttelarket4934]

[16^(1 - 1/2)]5^(1/(1/2) = 4*25 = 100.

[@Shobhamaths]

x=1/2
(2^4(1-x) )*5^1/x=2^2*5^2
Compasring 2 or 5 bas on both sides
4(1-x) =2
x=1/2

[@2012tulio]

1/2 is not the only solution but x= -1.161 also

[@Shobhamaths]

@2012tulio ok👍

[@2012tulio]

x = 0.5 and -1.161

[@gregevgeni1864]

x = 1/2

[@RajeshKumar-wu7ox]

X=1/2

[@Fjfurufjdfjd]

Λογαριθμιζω με βαση το 2εχω:(2χ-1)(log_2(5)+2χ)=0
2χ-1=0 , χ=1/2 ή χ=- log_2(5)/2

[@zawatsky]

Я что, один получил третий корень?🙄

[@Fjfurufjdfjd]

​@@zawatsky ποια ειναι και πώς τη βρηκες?

[@zawatsky]

@@Fjfurufjdfjd 16^(x-x²)*5=100^x. 100^x=(10)^2x=(5*2)^2x=5^2x*2^2x. 16^(x-x²)=2^[4(x-x²)]=2^(4x-4x²). 2^(4x-4x²)*5=5^2x*2^2x. 2^(4x-4x²-2x)*5^(1-2x)=2^(2x-4x²)*5^(1-2x)=1. Первый случай. 2x-4x²=0 ∧ 1-2x=0. x-2x²=0 ∧ x=½. ½-2*¼=½-2/4=½-½=0. x₁=½✔Второй случай. 5=4*1¼=2²*1¼=2^(2+lb1¼). 2^(2x-4x²)*2^(2+lb1¼)=2^(2x-4x²+2+lb1¼)=1⇒2x-4x²+2+lb1¼=0. x-2x²+1+lb1¼/2=x-2x²+1+lb(√5/2)=x-2x²+1+lb√5-lb2=x-2x²+1+lb√5-1=x-2x²+lb√5=0. 2x²-x-lb√5=0. D=1+4*2*lb√5=1+4lb5. √D=√(1+4lb5). x=[1±√(1+4lb5)]/4=¼±√[(1+4lb5)/16]=¼±√[1/16+(lb5)/4]. x₂=¼+√[1/16+(lb5)/4].✔ x₃=¼-√[1/16+(lb5)/4].✔

[@zawatsky]

16^(x-x²)*5=100^x. 100^x=(10)^2x=(5*2)^2x=5^2x*2^2x. 16^(x-x²)=2^[4(x-x²)]=2^(4x-4x²). 2^(4x-4x²)*5=5^2x*2^2x. 2^(4x-4x²-2x)*5^(1-2x)=2^(2x-4x²)*5^(1-2x)=1. Первый случай. 2x-4x²=0 ∧ 1-2x=0. x-2x²=0 ∧ x=½. ½-2*¼=½-2/4=½-½=0. x₁=½✔Второй случай. 5=4*1¼=2²*1¼=2^(2+lb1¼). 2^(2x-4x²)*2^(2+lb1¼)=2^(2x-4x²+2+lb1¼)=1⇒2x-4x²+2+lb1¼=0. x-2x²+1+lb1¼/2=x-2x²+1+lb(√5/2)=x-2x²+1+lb√5-lb2=x-2x²+1+lb√5-1=x-2x²+lb√5=0. 2x²-x-lb√5=0. D=1+4*2*lb√5=1+4lb5. √D=√(1+4lb5). x=[1±√(1+4lb5)]/4=¼±√[(1+4lb5)/16]=¼±√[1/16+(lb5)/4]. x₂=¼+√[1/16+(lb5)/4].✔ x₃=¼-√[1/16+(lb5)/4].✔

[@SimWeeKeat-c7e]

Do not be so complicated... just use normal comparison of power for indices.

[@walterwen2975]

MIND-BLOWING Algebra Secrets: [16^(1 - x)][5^(1/x)] = 100, x ≠ 0, x ϵ ℜ; x =?
log{[16^(1 - x)][5^(1/x)]} = log[16^(1 - x)] + log[5^(1/x)] = log100
(1 - x)log16 + (1/x)log5 = log100, (1 - x) + (1/x)log₁₆5 = log₁₆100; x ≠ 0
x - x² + log₁₆5 = (log₁₆100)x, x² + (log₁₆100 - 1)x - log₁₆5 = 0, x² + 0.661x - 0.581 = 0
x = {- 0.661 ± √[(- 0.661)² - 4(- 0.581)]}/2 = (- 0.661 ± √2.761)/2 = (- 0.661 ± 1.661)/2
x = (- 0.661 + 1.661)/2 = 1/2 or x = (- 0.661 - 1.661)/2 = - 1.161
Answer check:
x = 1/2, 1/x = 2: [16^(1 - x)][5^(1/x)] = (16¹⸍²)(5²) = (4)(25) = 100; Confirmed
x = - 1.161: (16²·¹⁶¹)(5⁻⁰·⁸⁶¹) = (400)(0.25) = 100; Confirmed
The calculation was achieved on a smartphone with a standard calculator app
Final answer:
x = 1/2 or x = - 1.161