What is the answer to log_{5}(x+4)+log_{5}(x+1)=1 ?

The answer to log_{5}(x+4)+log_{5}(x+1)=1 is x=(-5+sqrt(29))/2

Solve for x: log_{5}(x+4)+log_{5}(x+1)=1

The solution is x=(-5+sqrt(29))/2

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

x = \frac{- 5 + 29}{2}

Decimal Notation

x = 0.19258 \dots

Solution steps

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

Apply log rules

(x + 4) (x + 1) = 5

Solve

(x + 4) (x + 1) = 5 : x = \frac{- 5 + 29}{2}, x = \frac{- 5 - 29}{2}

x = \frac{- 5 + 29}{2}, x = \frac{- 5 - 29}{2}

Verify Solutions:

x = \frac{- 5 + 29}{2}

True

, x = \frac{- 5 - 29}{2}

False

The solution is

x = \frac{- 5 + 29}{2}

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

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

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

0 = - ln (x + 2)

lo g_{4} (x) - lo g_{x} (16) = 1

What is the answer to log_{5}(x+4)+log_{5}(x+1)=1 ?
The answer to log_{5}(x+4)+log_{5}(x+1)=1 is x=(-5+sqrt(29))/2
Solve for x: log_{5}(x+4)+log_{5}(x+1)=1
The solution is x=(-5+sqrt(29))/2