What is the answer to log_{x-1}(x^2-5x+7)=1 ?

The answer to log_{x-1}(x^2-5x+7)=1 is x=4

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log_{x-1}(x^2-5x+7)=1

lo g_{x - 1} (x^{2} - 5 x + 7) = 1

x = 4

Solution steps

lo g_{x - 1} (x^{2} - 5 x + 7) = 1

Apply log rules

\frac{ln ( x ^{2} - 5 x + 7 )}{ln ( x - 1 )} = 1

Multiply both sides by

ln (x - 1)

\frac{ln ( x ^{2} - 5 x + 7 )}{ln ( x - 1 )} ln (x - 1) = 1 \cdot ln (x - 1)

Simplify

ln (x^{2} - 5 x + 7) = ln (x - 1)

Apply log rules

x^{2} - 5 x + 7 = x - 1

Solve

x^{2} - 5 x + 7 = x - 1 : x = 4, x = 2

x = 4, x = 2

Verify Solutions:

x = 4

True

, x = 2

False

The solution is

x = 4

lo g_{10} (25 x) = 3

lo g_{3} (3 - x) = 3

lo g_{x} (0.0001) = - 4

lo g_{x} (32) = \frac{5}{3}

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

What is the answer to log_{x-1}(x^2-5x+7)=1 ?
The answer to log_{x-1}(x^2-5x+7)=1 is x=4
Solve for x: log_{x-1}(x^2-5x+7)=1
The solution is x=4