What is the answer to log_{x}(x^2-5x-4)=1 ?

The answer to log_{x}(x^2-5x-4)=1 is x=3+sqrt(13)

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log_{x}(x^2-5x-4)=1

lo g_{x} (x^{2} - 5 x - 4) = 1

x = 3 + 13

Solution steps

lo g_{x} (x^{2} - 5 x - 4) = 1

Apply log rules

\frac{ln ( x ^{2} - 5 x - 4 )}{ln ( x )} = 1

Multiply both sides by

ln (x)

\frac{ln ( x ^{2} - 5 x - 4 )}{ln ( x )} ln (x) = 1 \cdot ln (x)

Simplify

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

Apply log rules

x^{2} - 5 x - 4 = x

Solve

x^{2} - 5 x - 4 = x : x = 3 + 13, x = 3 - 13

x = 3 + 13, x = 3 - 13

Verify Solutions:

x = 3 + 13

True

, x = 3 - 13

False

The solution is

x = 3 + 13

ln (4 - c) = 0

lo g_{10} (7 x + 3) = 2

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

ln (x) + ln (x^{3}) = 0

ln (x - 6) + ln (x) = ln (x - 30)

What is the answer to log_{x}(x^2-5x-4)=1 ?
The answer to log_{x}(x^2-5x-4)=1 is x=3+sqrt(13)
Solve for x: log_{x}(x^2-5x-4)=1
The solution is x=3+sqrt(13)