What is the answer to log_{3}(x+3)+log_{3}(4x+7)=2 ?

The answer to log_{3}(x+3)+log_{3}(4x+7)=2 is x=-3/4

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log_{3}(x+3)+log_{3}(4x+7)=2

lo g_{3} (x + 3) + lo g_{3} (4 x + 7) = 2

x = - \frac{3}{4}

Decimal Notation

x = - 0.75

Solution steps

lo g_{3} (x + 3) + lo g_{3} (4 x + 7) = 2

Apply log rules

(x + 3) (4 x + 7) = 9

Solve

(x + 3) (4 x + 7) = 9 : x = - \frac{3}{4}, x = - 4

x = - \frac{3}{4}, x = - 4

Verify Solutions:

x = - \frac{3}{4}

True

, x = - 4

False

The solution is

x = - \frac{3}{4}

y, x = ln (2 y + 1)

ln (x + 9) - ln (x + 7) = 3

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

lo g_{3} (\frac{1}{x}) = - 2

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

What is the answer to log_{3}(x+3)+log_{3}(4x+7)=2 ?
The answer to log_{3}(x+3)+log_{3}(4x+7)=2 is x=-3/4
Solve for x: log_{3}(x+3)+log_{3}(4x+7)=2
The solution is x=-3/4