What is the answer to log_{5}(2x+1)=1-log_{5}(x+2) ?

The answer to log_{5}(2x+1)=1-log_{5}(x+2) is x= 1/2

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log_{5}(2x+1)=1-log_{5}(x+2)

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

x = \frac{1}{2}

Decimal Notation

x = 0.5

Solution steps

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

Add

lo g_{5} (x + 2)

to both sides

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

Simplify

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

Apply log rules

(2 x + 1) (x + 2) = 5

Solve

(2 x + 1) (x + 2) = 5 : x = \frac{1}{2}, x = - 3

x = \frac{1}{2}, x = - 3

Verify Solutions:

x = \frac{1}{2}

True

, x = - 3

False

The solution is

x = \frac{1}{2}

ln (y) = x + ln (x), y

lo g_{5} (x + 4) + lo g_{5} (x + 1) = 1

lo g_{5} (x + 4) + lo g_{5} (x + 1) = 2

2 ln (\frac{x}{2}) = - e

x, 6 lo g_{10} (x) = 9

What is the answer to log_{5}(2x+1)=1-log_{5}(x+2) ?
The answer to log_{5}(2x+1)=1-log_{5}(x+2) is x= 1/2
Solve for x: log_{5}(2x+1)=1-log_{5}(x+2)
The solution is x= 1/2