What is the answer to log_{3+x}(x^2-x)=1 ?

The answer to log_{3+x}(x^2-x)=1 is x=3,x=-1

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log_{3+x}(x^2-x)=1

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

x = 3, x = - 1

Solution steps

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

Apply log rules

\frac{ln ( x ^{2} - x )}{ln ( 3 + x )} = 1

Multiply both sides by

ln (3 + x)

\frac{ln ( x ^{2} - x )}{ln ( 3 + x )} ln (3 + x) = 1 \cdot ln (3 + x)

Simplify

ln (x^{2} - x) = ln (3 + x)

Apply log rules

x^{2} - x = 3 + x

Solve

x^{2} - x = 3 + x : x = 3, x = - 1

x = 3, x = - 1

Verify Solutions:

x = 3

True

, x = - 1

True

The solutions are

x = 3, x = - 1

lo g_{2} (4 x - 9) = 1

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

x, ln (y) + y = x + ln (x) + c

lo g_{2} (3 x - 2) = 1 + lo g_{2} (5)

x, ln (x) = - 2.27

What is the answer to log_{3+x}(x^2-x)=1 ?
The answer to log_{3+x}(x^2-x)=1 is x=3,x=-1
Solve for x: log_{3+x}(x^2-x)=1
The solution is x=3,x=-1