What is the answer to log_{3}(x-4)+log_{3}(x+6)=1 ?

The answer to log_{3}(x-4)+log_{3}(x+6)=1 is x=-1+2sqrt(7)

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log_{3}(x-4)+log_{3}(x+6)=1

lo g_{3} (x - 4) + lo g_{3} (x + 6) = 1

x = - 1 + 27

Decimal Notation

x = 4.29150 \dots

Solution steps

lo g_{3} (x - 4) + lo g_{3} (x + 6) = 1

Apply log rules

(x - 4) (x + 6) = 3

Solve

(x - 4) (x + 6) = 3 : x = - 1 + 27, x = - 1 - 27

x = - 1 + 27, x = - 1 - 27

Verify Solutions:

x = - 1 + 27

True

, x = - 1 - 27

False

The solution is

x = - 1 + 27

lo g_{x} (1.41) = 0.5

- 3 = 20 lo g_{10} (x)

lo g_{5} (x + 1) = 2 + lo g_{5} (x - 1)

e^{2} ln (b) = 400

x = 4 lo g_{10} (x)

What is the answer to log_{3}(x-4)+log_{3}(x+6)=1 ?
The answer to log_{3}(x-4)+log_{3}(x+6)=1 is x=-1+2sqrt(7)
Solve for x: log_{3}(x-4)+log_{3}(x+6)=1
The solution is x=-1+2sqrt(7)