What is the answer to log_{3}(2x+3)=4-log_{3}(x+6) ?

The answer to log_{3}(2x+3)=4-log_{3}(x+6) is x=3

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log_{3}(2x+3)=4-log_{3}(x+6)

lo g_{3} (2 x + 3) = 4 - lo g_{3} (x + 6)

x = 3

Solution steps

lo g_{3} (2 x + 3) = 4 - lo g_{3} (x + 6)

Add

lo g_{3} (x + 6)

to both sides

lo g_{3} (2 x + 3) + lo g_{3} (x + 6) = 4 - lo g_{3} (x + 6) + lo g_{3} (x + 6)

Simplify

lo g_{3} (2 x + 3) + lo g_{3} (x + 6) = 4

Apply log rules

(2 x + 3) (x + 6) = 81

Solve

(2 x + 3) (x + 6) = 81 : x = 3, x = - \frac{21}{2}

x = 3, x = - \frac{21}{2}

Verify Solutions:

x = 3

True

, x = - \frac{21}{2}

False

The solution is

x = 3

lo g_{3} (2) + lo g_{3} (b + 1) = 1

3^{l o g_{x} (81)} = x

lo g_{x} (625) = - 4

lo g_{9} (2 x) - lo g_{3} (2) = 1

lo g_{3} (4 x - 2) = 4

What is the answer to log_{3}(2x+3)=4-log_{3}(x+6) ?
The answer to log_{3}(2x+3)=4-log_{3}(x+6) is x=3
Solve for x: log_{3}(2x+3)=4-log_{3}(x+6)
The solution is x=3