What is the answer to (log_{e}(x-1))^2=4 ?

The answer to (log_{e}(x-1))^2=4 is x=e^2+1,x= 1/(e^2)+1

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(log_{e}(x-1))^2=4

(lo g_{e} (x - 1))^{2} = 4

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

Decimal Notation

x = 8.38905 \dots, x = 1.13533 \dots

Solution steps

(lo g_{e} (x - 1))^{2} = 4

Rewrite the equation with

lo g_{e} (x - 1) = u

(u)^{2} = 4

Solve

u^{2} = 4 : u = 4, u = - 4

u = 4, u = - 4

Substitute back

u = lo g_{e} (x - 1),

solve for

x

Solve

lo g_{e} (x - 1) = 4 : x = e^{2} + 1

Solve

lo g_{e} (x - 1) = - 4 : x = \frac{1}{e ^{2}} + 1

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

Verify Solutions:

x = e^{2} + 1

True

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

True

The solutions are

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

x ln (2) + ln (5) = ln (4^{x} - 6)

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

ln (x) + ln (3) = ln (9)

lo g_{6} (3 - m) + lo g_{6} (- m - 2) = 2

lo g_{6} (3 x) = 2

What is the answer to (log_{e}(x-1))^2=4 ?
The answer to (log_{e}(x-1))^2=4 is x=e^2+1,x= 1/(e^2)+1