What is the answer to 5402=-ln(1-x) ?

The answer to 5402=-ln(1-x) is No Solution for x\in\mathbb{R}

Solve for x: 5402=-ln(1-x)

The solution is No Solution for x\in\mathbb{R}

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5402 = - ln (1 - x)

No Solution for x \in R

Solution steps

5402 = - ln (1 - x)

Move

ln (1 - x)

to the left side

5402 + ln (1 - x) = 0

Move

5402

to the right side

ln (1 - x) = - 5402

Apply log rules

1 - x = \frac{1}{e ^{5402}}

Solve

1 - x = \frac{1}{e ^{5402}} : x = - \frac{1 - e ^{5402}}{e ^{5402}}

x = - \frac{1 - e ^{5402}}{e ^{5402}}

Verify Solutions:

x = - \frac{1 - e ^{5402}}{e ^{5402}}

False

The solution is

No Solution for x \in R

x, 3 = lo g_{10} (x)

lo g_{x} (x + 20) = 2

lo g_{4} (x - 95) = lo g_{4} (x - 5) + 2

lo g_{10} (100 x) = 3

lo g_{2} (x + 4) = 1 + lo g_{2} (x + 3)

What is the answer to 5402=-ln(1-x) ?
The answer to 5402=-ln(1-x) is No Solution for x\in\mathbb{R}
Solve for x: 5402=-ln(1-x)
The solution is No Solution for x\in\mathbb{R}