Question

1.Determine log x base 2401 when X^3 =7
2.2x+4 squared = 1+ square X+9. Find x
3. Expand to the fourth term the expression (x^2- y^2 + 2)^1/2

Answer 1

logarithms

$x^3=7\\\\Log_7\ x^3=Log_7\ 7=1\\\\3\ Log_7\ x=1,\ \ Log_7\ x=1/3\\\\let\ y=Log_{2401}\ x\\\\2401^y=x\\\\. [\ (49)^2\ ]^y=x\\\\Log_7\ (49)^{2y}=Log_7\ x\\\\2y\ Log_7\ 49=Log_7\ x\\\\2y\ *\ 2=1/3,\ calculated\ above\\\\y=1/12$

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(2x + 4)²  =  1 +  x² + 9 ???
question is  not clear ??
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3.
       √(x² - y² + 2)  =  ?  the taylor series ?

  expansion about point  (x0, y0) = (0, 0).

$F= \sqrt{x^2-y^2+2}\\\\\frac{dF}{dx}=\frac{2x}{2\sqrt{x^2-y^2+2}},\ \ at\ (0,0):\ 0\\\\\frac{dF}{dy}=\frac{-2y}{2\sqrt{x^2-y^2+2}},\ \ value\ at\ (0,0):\ 0\\\\\frac{d^2F}{dx\ dy}=\frac{xy}{(x^2-y^2+2)^\frac{3}{2}},\ \ va;ue\ at\ (0,0),\ =0\\\\\frac{d^2F}{dx^2}=\frac{-2x^2}{2(x^2-y^2+2)^\frac{3}{2}}+\frac{1}{\sqrt{x^2-y^2+2}}=\frac{2-y^2}{(x^2-y^2+2)^\frac{3}{2}},\ value\ at\ (0,0)\ :\frac{1}{\sqrt2}$

$\frac{d^2F}{dy^2}=\frac{y^2}{(x^2-y^2+2)^\frac{3}{2}}-\frac{1}{\sqrt{x^2-y^2+2}}=\frac{2y^2-x^2-2}{(x^2-y^2+2)^\frac{3}{2}},\ \ value\ at\ (0,0)\ :\ \frac{-1}{\sqrt2}$

Answer 2

Answer:

Step-by-step explanation: