What is the answer to log_{3}(2x)+log_{3}(x-1)=1 ?

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

Solve for x: log_{3}(2x)+log_{3}(x-1)=1

The solution is x=(1+sqrt(7))/2

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

x = \frac{1 + 7}{2}

Decimal Notation

x = 1.82287 \dots

Solution steps

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

Apply log rules

2 x (x - 1) = 3

Solve

2 x (x - 1) = 3 : x = \frac{1 + 7}{2}, x = \frac{1 - 7}{2}

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

Verify Solutions:

x = \frac{1 + 7}{2}

True

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

False

The solution is

x = \frac{1 + 7}{2}

2.44 = - lo g_{10} (x)

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

(q) = 5 + 10 lo g_{10} (10 + 2 q)

lo g_{10} (2 x) = 1

x, 5 ln (4 x + 4) + 4 = - 11

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