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

The answer to log_{7}(2x+1)+log_{7}(x-1)=1 is x=(1+sqrt(65))/4

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

The solution is x=(1+sqrt(65))/4

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

x = \frac{1 + 65}{4}

Solution steps

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

Apply log rules

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

Solve

(2 x + 1) (x - 1) = 7 : x = \frac{1 + 65}{4}, x = \frac{1 - 65}{4}

x = \frac{1 + 65}{4}, x = \frac{1 - 65}{4}

Verify Solutions:

x = \frac{1 + 65}{4}

True

, x = \frac{1 - 65}{4}

False

The solution is

x = \frac{1 + 65}{4}

ln (e - 8 x + 1 - 1) = 20, x

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

2 ln (t + 2) = 0

lo g_{10} (a x) = 6

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

What is the answer to log_{7}(2x+1)+log_{7}(x-1)=1 ?
The answer to log_{7}(2x+1)+log_{7}(x-1)=1 is x=(1+sqrt(65))/4
Solve for x: log_{7}(2x+1)+log_{7}(x-1)=1
The solution is x=(1+sqrt(65))/4