What is the answer to log_{x-1}(4+3x-x^2)=2 ?

The answer to log_{x-1}(4+3x-x^2)=2 is x=3

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log_{x-1}(4+3x-x^2)=2

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

x = 3

Solution steps

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

Apply log rules

\frac{ln ( 4 + 3 x - x ^{2} )}{ln ( x - 1 )} = 2

Multiply both sides by

ln (x - 1)

\frac{ln ( 4 + 3 x - x ^{2} )}{ln ( x - 1 )} ln (x - 1) = 2 ln (x - 1)

Simplify

ln (4 + 3 x - x^{2}) = 2 ln (x - 1)

Apply log rules

4 + 3 x - x^{2} = (x - 1)^{2}

Solve

4 + 3 x - x^{2} = (x - 1)^{2} : x = - \frac{1}{2}, x = 3

x = - \frac{1}{2}, x = 3

Verify Solutions:

x = - \frac{1}{2}

False

, x = 3

True

The solution is

x = 3

ln (x) + ln (x + 1) = 5

ln (x - 3) - ln (x - 5) = ln (5)

ln (x) + 2 = 0

lo g_{2 x + 3} (81) = 2

ln (2 x - 3) - ln (x + 1) = e

What is the answer to log_{x-1}(4+3x-x^2)=2 ?
The answer to log_{x-1}(4+3x-x^2)=2 is x=3
Solve for x: log_{x-1}(4+3x-x^2)=2
The solution is x=3