Alan

No they are the same.  The roots don't have to be different! You can have repeated roots.

There are 6 possible solutions to (cos(60)+sin(60)*i)⁶ = 1:

If, on the other hand, you meant  (1/5)^(4x-3) = 5 then:

5^(3 - 4x) = 5  so

3 - 4x = 1

2 = 4x

x = 1/2

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-3^2 = -9   because exponentiation takes priority over negation.

(-3)^2 = +9

24 = 150/(5 + 25e^-0.45x)

24 = 30/(1 + 5e^-0.45x)

1 + 5e^-0.45x = 30/24

1 + 5e^-0.45x = 5/4

5e^-0.45x = 1/4

e^-0.45x = 1/20

-0.45x = ln(1/20)

0.45x = ln(20)

x = ln(20)/0.45 

x ≈ 6.657

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"completing the square" is the correct term in English.

5ax² - 10ax + 7 = 0

Divide through by 5a

x² - 2x + 7/(5a) = 0

Add 1 to both sides:

x² - 2x + 1 + 7/(5a) = 1

(x - 1)² +7/(5a) = 1

Subtract 7/(5a) from both sides

(x - 1)² = 1 - 7/(5a)

x - 1 = sqrt(1 - 7/(5a))   and  x - 1 = -sqrt(1 - 7/(5a))

x = 1 + sqrt(1 - 7/(5a))  and  x = 1 - sqrt(1 - 7/(5a))

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Let z = 3^h

lnz = h*ln3

(1/z)dz/dh = ln3

dz/dh = ln3*z → ln3*3^h

d3^h/dh = ln3*3^h

Taylor series of function f(h):  f(h) = f(0) + df/dh₀*h + (d²f/dh²)₀*h²/2! + ...

If f(h) = 3^h then df/dh = ln3*3^h,  d²f/dh² = (ln3)²*3^h, etc. 

So, noting that 3⁰ = 1, we have:  3^h = 1 + ln3*h + (ln3)²*h²/2! + ...

(the ... indicates terms involving higher powers of h)

Hence 3^h - 1 = ln3*h + (ln3)²*h²/2! + ...

(3^h - 1)/h = ln3 + (ln3)²*h/2! + ...

As h → 0 all but the first term on the right hand side go to 0, so in the limit we have

lim_h→0(3^h - 1)/h = ln3 

(much easier using l'Hopital's rule though!)

The relative speed of the trains is 75+65 → 140 mph, so this is the same as asking how long does it take a train travelling at 140mph to travel 196miles.  This is 196/140 hours or 1.4 hours or 1 hour 24 minutes.

cos⁴x - sin⁴x → (cos²x + sin²x)(cos²x - sin²x) → 1*(cos²x - sin²x) → cos(2x)

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