What is the answer to ln(x-2)=in^{12}-in(x+2) ?

The answer to ln(x-2)=in^{12}-in(x+2) is x=(\W(\frac{in)/(e^{-in^{12)+4in}})+2in}{in}

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ln(x-2)=in^{12}-in(x+2)

ln (x - 2) = i n^{12} - in (x + 2)

x = \frac{W ( \frac{in}{e ^{- i n^{12} + 4 in}} ) + 2 in}{in}

Solution steps

ln (x - 2) = i n^{12} - in (x + 2)

f (x) = g (x),

then

a^{f (x)} = a^{g (x)}

e^{l n (x - 2)} = e^{i n^{12} - in (x + 2)}

Simplify

e^{l n (x - 2)} : x - 2

x - 2 = e^{i n^{12} - in (x + 2)}

Solve

x - 2 = e^{i n^{12} - in (x + 2)} : x = \frac{W ( \frac{in}{e ^{- i n^{12} + 4 in}} ) + 2 in}{in}

x = \frac{W ( \frac{in}{e ^{- i n^{12} + 4 in}} ) + 2 in}{in}

lo g_{3} ((x + 1) (x + 3)) = 1

lo g_{2} (lo g_{e} (x)) = 1

lo g_{e} (- \frac{x}{456}) = 1

lo g_{10} (v) = \frac{4.288}{2.3}

lo g_{10} (15 - 2) = 2 lo g_{10} (x)

What is the answer to ln(x-2)=in^{12}-in(x+2) ?
The answer to ln(x-2)=in^{12}-in(x+2) is x=(\W(\frac{in)/(e^{-in^{12)+4in}})+2in}{in}