What is the answer to 0=ln(x-1)+2 ?

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

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0=ln(x-1)+2

0 = ln (x - 1) + 2

x = \frac{1}{e ^{2}} + 1

Decimal Notation

x = 1.13533 \dots

Solution steps

0 = ln (x - 1) + 2

Switch sides

ln (x - 1) + 2 = 0

Move

2

to the right side

ln (x - 1) = - 2

Apply log rules

x - 1 = \frac{1}{e ^{2}}

Solve

x - 1 = \frac{1}{e ^{2}} : x = \frac{1}{e ^{2}} + 1

x = \frac{1}{e ^{2}} + 1

Verify Solutions:

x = \frac{1}{e ^{2}} + 1

True

The solution is

x = \frac{1}{e ^{2}} + 1

0 = ln (x - 1) - 2

0.367 = 2 - lo g_{10} (x)

lo g_{a} (64) = - 2

x + lo g_{10} (x + 4) = 6

lo g_{2} (9 x + 12) - lo g_{2} (3) = 4