What is the answer to n(x)=log_{2/e}(x) ?

The answer to n(x)=log_{2/e}(x) is x=-(\W(-(ln(2)-1)n))/((ln(2)-1)n)

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n(x)=log_{2/e}(x)

n (x) = lo g_{\frac{2}{e}} (x)

x = - \frac{W ( - ( ln ( 2 ) - 1 ) n )}{( ln ( 2 ) - 1 ) n}

Solution steps

n (x) = lo g_{\frac{2}{e}} (x)

f (x) = g (x),

then

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

(\frac{2}{e})^{n x} = (\frac{2}{e})^{l o g_{\frac{2}{e}} (x)}

Simplify

(\frac{2}{e})^{l o g_{\frac{2}{e}} (x)} : x

(\frac{2}{e})^{n x} = x

Solve

(\frac{2}{e})^{n x} = x : x = - \frac{W ( - ( ln ( 2 ) - 1 ) n )}{( ln ( 2 ) - 1 ) n}

x = - \frac{W ( - ( ln ( 2 ) - 1 ) n )}{( ln ( 2 ) - 1 ) n}

Verify Solutions:

x = - \frac{W ( - ( ln ( 2 ) - 1 ) n )}{( ln ( 2 ) - 1 ) n}

True

The solution is

x = - \frac{W ( - ( ln ( 2 ) - 1 ) n )}{( ln ( 2 ) - 1 ) n}

ln (x) = \frac{90}{10}

lo g_{2} (1 + a) = 4

335 = lo g_{93} (x)

ln (16 x) = ln (16) + 4 ln (2)

lo g_{3} (\frac{x + 3}{2}) = 2

What is the answer to n(x)=log_{2/e}(x) ?
The answer to n(x)=log_{2/e}(x) is x=-(\W(-(ln(2)-1)n))/((ln(2)-1)n)