What is the answer to ((2e)^0)/(ln(2e))+C= 1/(ln(2)+1) ?

The answer to ((2e)^0)/(ln(2e))+C= 1/(ln(2)+1) is C= 1/(ln(2)+1)-1/(ln(2e))

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((2e)^0)/(ln(2e))+C= 1/(ln(2)+1)

\frac{( 2 e ) ^{0}}{ln ( 2 e )} + C = \frac{1}{ln ( 2 ) + 1}

C = \frac{1}{ln ( 2 ) + 1} - \frac{1}{ln ( 2 e )}

Decimal

C = 0

Solution steps

\frac{( 2 e ) ^{0}}{ln ( 2 e )} + C = \frac{1}{ln ( 2 ) + 1}

Apply rule

a^{0} = 1, a \neq = 0

(2 e)^{0} = 1

\frac{1}{ln ( 2 e )} + C = \frac{1}{ln ( 2 ) + 1}

Move

\frac{1}{ln ( 2 e )}

to the right side

C = \frac{1}{ln ( 2 ) + 1} - \frac{1}{ln ( 2 e )}

f, f = {- 2}

(x + 30) + (x + 25) = 7 x

\frac{v + 25}{4} = 9

x, 3 x + 4 y = 5

\frac{x - 3}{2} - \frac{x - 2}{3} = \frac{x + 5}{4}

What is the answer to ((2e)^0)/(ln(2e))+C= 1/(ln(2)+1) ?
The answer to ((2e)^0)/(ln(2e))+C= 1/(ln(2)+1) is C= 1/(ln(2)+1)-1/(ln(2e))