What is the answer to log_{x+1}((x^2+x-6)^2)=4 ?

The answer to log_{x+1}((x^2+x-6)^2)=4 is x=1

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log_{x+1}((x^2+x-6)^2)=4

lo g_{x + 1} ((x^{2} + x - 6)^{2}) = 4

x = 1

Solution steps

lo g_{x + 1} ((x^{2} + x - 6)^{2}) = 4

Apply log rules

\frac{ln ( ( x ^{2} + x - 6 ) ^{2} )}{ln ( x + 1 )} = 4

Multiply both sides by

ln (x + 1)

\frac{ln ( ( x ^{2} + x - 6 ) ^{2} )}{ln ( x + 1 )} ln (x + 1) = 4 ln (x + 1)

Simplify

ln ((x^{2} + x - 6)^{2}) = 4 ln (x + 1)

Apply log rules

(x^{2} + x - 6)^{2} = (x + 1)^{4}

Solve

(x^{2} + x - 6)^{2} = (x + 1)^{4} : x = 1, x = - \frac{5}{2}, x = - 7

x = 1, x = - \frac{5}{2}, x = - 7

Verify Solutions:

x = 1

True

, x = - \frac{5}{2}

False

, x = - 7

False

The solution is

x = 1

3 lo g_{10} (8 t + 10) = 0

ln (x) = 6.666

y, ln (y) = 2 ln (x)

lo g_{3} (x + 3) = 0

lo g_{3} (x + 15) - lo g_{3} (x - 1) = 2

What is the answer to log_{x+1}((x^2+x-6)^2)=4 ?
The answer to log_{x+1}((x^2+x-6)^2)=4 is x=1
Solve for x: log_{x+1}((x^2+x-6)^2)=4
The solution is x=1