What is the answer to log_{x}(x^2+x-6)=1 ?

The answer to log_{x}(x^2+x-6)=1 is x=sqrt(6)

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log_{x}(x^2+x-6)=1

lo g_{x} (x^{2} + x - 6) = 1

x = 6

Solution steps

lo g_{x} (x^{2} + x - 6) = 1

Apply log rules

\frac{ln ( x ^{2} + x - 6 )}{ln ( x )} = 1

Multiply both sides by

ln (x)

\frac{ln ( x ^{2} + x - 6 )}{ln ( x )} ln (x) = 1 \cdot ln (x)

Simplify

ln (x^{2} + x - 6) = ln (x)

Apply log rules

x^{2} + x - 6 = x

Solve

x^{2} + x - 6 = x : x = 6, x = - 6

x = 6, x = - 6

Verify Solutions:

x = 6

True

, x = - 6

False

The solution is

x = 6

10 lo g_{10} (x) = 0.17

40 = 6 \cdot lo g_{2} (x)

x = 9 + ln (x)

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

x, lo g_{10} (x + 1) = 5

What is the answer to log_{x}(x^2+x-6)=1 ?
The answer to log_{x}(x^2+x-6)=1 is x=sqrt(6)
Solve for x: log_{x}(x^2+x-6)=1
The solution is x=sqrt(6)