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

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

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

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

x = \frac{1}{2}

Decimal Notation

x = 0.5

Solution steps

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

Apply log rules

\frac{ln ( 2 x + 1 )}{ln ( x )} = - 1

Multiply both sides by

ln (x)

\frac{ln ( 2 x + 1 )}{ln ( x )} ln (x) = - 1 \cdot ln (x)

Simplify

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

Add

ln (x)

to both sides

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

Simplify

ln (2 x + 1) + ln (x) = 0

Apply log rules

(2 x + 1) x = 1

Solve

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

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

Verify Solutions:

x = \frac{1}{2}

True

, x = - 1

False

The solution is

x = \frac{1}{2}

x, ln (x + y - 5) = 0

lo g_{4} (x) - lo g_{8} (x) = 1

2 ln (x) = 14

lo g_{10} (x) = 2634166

x, lo g_{p} (x) = - 3

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