What is the answer to solvefor x,d^3+y=e^{-x}+1 ?

The answer to solvefor x,d^3+y=e^{-x}+1 is x=-ln(d^3+y-1)

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solvefor x,d^3+y=e^{-x}+1

solve for

x, d^{3} + y = e^{- x} + 1

x = - ln (d^{3} + y - 1)

Solution steps

d^{3} + y = e^{- x} + 1

Switch sides

e^{- x} + 1 = d^{3} + y

Move

1

to the right side

e^{- x} = d^{3} + y - 1

Apply exponent rules

- x = ln (d^{3} + y - 1)

Solve

- x = ln (d^{3} + y - 1) : x = - ln (d^{3} + y - 1)

x = - ln (d^{3} + y - 1)

100 0^{m} = 100000

\frac{( 5 ^{x + 1} \cdot 62 5 ^{3} )}{25} = 5^{x}

2^{x} = 2^{2 x} + 1

9^{x} + 8 \cdot 3^{x} = 9

\frac{x}{7 ^{3 + x}} = 1

What is the answer to solvefor x,d^3+y=e^{-x}+1 ?
The answer to solvefor x,d^3+y=e^{-x}+1 is x=-ln(d^3+y-1)