What is the answer to log_{7}(2x+1)+log_{7}(x-2)=1 ?

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

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log_{7}(2x+1)+log_{7}(x-2)=1

lo g_{7} (2 x + 1) + lo g_{7} (x - 2) = 1

x = 3

Solution steps

lo g_{7} (2 x + 1) + lo g_{7} (x - 2) = 1

Apply log rules

(2 x + 1) (x - 2) = 7

Solve

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

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

Verify Solutions:

x = 3

True

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

False

The solution is

x = 3

ln (x^{2} + 1) + 1 = ln (x + 1)

lo g_{3} (6 x + 5) = 2

lo g_{10} (4 x) + lo g_{10} (13) = 0

y, x = - lo g_{e} (y)

ln (2 x - 1) = ln (3 x)

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