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

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

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

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

x = 1, x = - 3

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 = 1, x = - 3

x = 1, x = - 3

Verify Solutions:

x = 1

True

, x = - 3

True

The solutions are

x = 1, x = - 3

1 - 2 lo g_{10} (x) = - 4

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

lo g_{2} (4 x + 5) = lo g_{2} (2 x + 11)

5 lo g_{5} (2) = lo g_{5} (x)

ln (p^{2} - p) = ln (6 p + 18)

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