We have 8^x+2^x=130. 230 can be written as 128+2. So the equation is 8^x+2^x=128+2. 8^xis 2^3x+xwhich means 2^4x. So 2^4x=2^7+1. So we have 4x=8.Hence xis 2.
2^×)^3+2^×=5^3+5 =>(2^×-5)+{(2^×)^3-5^5)}=0 =>(2^×-5)+(2^×-5){(2^2×+5(2^×)+5^2)}=0 =>Either 2^×=5.... =>×=log@2 (5) Or (2^×)^2+5(2^×)+26=0.. 2^× ={-5+_ _/5^2-4(26)}/2 Which goes for a complex soln.
8^x = (2^3)^x = (2^x)^3. Let 2^x = u. u^3 + u - 130 = 0. Solve then back substitute. u = 5. 5 = 2^x --> ln(5)/ln(2) for the real solution. Quadratic on x^2 + 5x + 26 for the 2 complex solutions.
Even you can use Vieta's Formula frim 5:30 onwards to find the roots directly. No need to follow the lond D=b^2-4ca.
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We have 8^x+2^x=130.
230 can be written as 128+2. So the equation is 8^x+2^x=128+2.
8^xis 2^3x+xwhich means 2^4x. So 2^4x=2^7+1. So we have 4x=8.Hence xis 2.
2^×)^3+2^×=5^3+5
=>(2^×-5)+{(2^×)^3-5^5)}=0
=>(2^×-5)+(2^×-5){(2^2×+5(2^×)+5^2)}=0
=>Either 2^×=5....
=>×=log@2 (5)
Or (2^×)^2+5(2^×)+26=0..
2^×
={-5+_ _/5^2-4(26)}/2
Which goes for a complex soln.
8^x = (2^3)^x = (2^x)^3. Let 2^x = u. u^3 + u - 130 = 0. Solve then back substitute. u = 5. 5 = 2^x --> ln(5)/ln(2) for the real solution. Quadratic on x^2 + 5x + 26 for the 2 complex solutions.
Kul
Answer 2.322
130= 5^3+5= m^3+m
Some standard problems should be uploaded
Why would 1(m-5) transform into just 1? Where does (m-5) go?
За скобки вынес
The m-5 has been factored outside the bracket
How is m^3 - 5^3 = (m-5)(m^2 +5m + 5^2)??????
(m-5)(m^2 +5m + 5^2)= m^3 + 5•m^2 - 25m - 5^3 😮😮 please explain
(m-5)(m^2 + 5m + 5^2) = m^3 + 5m^2 +25m - 5m^2 - 25m - 5^3
log5/log2
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If x=2, it's 64 + 4 = 68 (too small)
If x=3, it's 64(8) + 8 = 512 + 8 = 520 (too big)
If x=2.5, it's 186.68 (still too big, but better
If x=2.25, it's 112.39 (better, but slightly lesser) 🤔
If x=2.3, it's 124.35 (almost there)
If x=2.3225, it's 130.15. Close enough tbh.🤔
this is not how math works
haha the trick which always works
স্যারআমিগণিতপারিনা😢😢
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