What is the answer to -3-ln(-3+x)=0 ?

The answer to -3-ln(-3+x)=0 is x= 1/(e^3)+3

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-3-ln(-3+x)=0

- 3 - ln (- 3 + x) = 0

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

Solution steps

- 3 - ln (- 3 + x) = 0

Move

3

to the right side

- ln (- 3 + x) = 3

Divide both sides by

- 1

ln (- 3 + x) = - 3

Apply log rules

- 3 + x = \frac{1}{e ^{3}}

Solve

- 3 + x = \frac{1}{e ^{3}} : x = \frac{1}{e ^{3}} + 3

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

Verify Solutions:

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

True

The solution is

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

a^{5} = lo g_{5} (x^{5})

lo g_{X} (16) = - 1

ln (5 x - 4) = ln (8) - ln (x - 2)

lo g_{10} (x) = \frac{- 1}{n}

ln (\frac{x ^{2}}{x + 2}) = 0

What is the answer to -3-ln(-3+x)=0 ?
The answer to -3-ln(-3+x)=0 is x= 1/(e^3)+3
Solve for x: -3-ln(-3+x)=0
The solution is x= 1/(e^3)+3