What is the answer to log_{x+6}(2x^2+7x)=2 ?

The answer to log_{x+6}(2x^2+7x)=2 is x=-4,x=9

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log_{x+6}(2x^2+7x)=2

lo g_{x + 6} (2 x^{2} + 7 x) = 2

x = - 4, x = 9

Solution steps

lo g_{x + 6} (2 x^{2} + 7 x) = 2

Apply log rules

\frac{ln ( 2 x ^{2} + 7 x )}{ln ( x + 6 )} = 2

Multiply both sides by

ln (x + 6)

\frac{ln ( 2 x ^{2} + 7 x )}{ln ( x + 6 )} ln (x + 6) = 2 ln (x + 6)

Simplify

ln (2 x^{2} + 7 x) = 2 ln (x + 6)

Apply log rules

2 x^{2} + 7 x = (x + 6)^{2}

Solve

2 x^{2} + 7 x = (x + 6)^{2} : x = - 4, x = 9

x = - 4, x = 9

Verify Solutions:

x = - 4

True

, x = 9

True

The solutions are

x = - 4, x = 9

ln (x + 2) - ln (x - 6) = 1

\frac{x}{ln ( x )} = 0

lo g_{10} (x) 64 = 3

ln (x) + ln (x + 1) = 3

0 = 2 ln (x - 2)

What is the answer to log_{x+6}(2x^2+7x)=2 ?
The answer to log_{x+6}(2x^2+7x)=2 is x=-4,x=9
Solve for x: log_{x+6}(2x^2+7x)=2
The solution is x=-4,x=9