What is the answer to log_{2}(x+1)+log_{2}(x-2)=3 ?

The answer to log_{2}(x+1)+log_{2}(x-2)=3 is x=(1+sqrt(41))/2

Solve for x: log_{2}(x+1)+log_{2}(x-2)=3

The solution is x=(1+sqrt(41))/2

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lo g_{2} (x + 1) + lo g_{2} (x - 2) = 3

x = \frac{1 + 41}{2}

Decimal Notation

x = 3.70156 \dots

Solution steps

lo g_{2} (x + 1) + lo g_{2} (x - 2) = 3

Apply log rules

(x + 1) (x - 2) = 8

Solve

(x + 1) (x - 2) = 8 : x = \frac{1 + 41}{2}, x = \frac{1 - 41}{2}

x = \frac{1 + 41}{2}, x = \frac{1 - 41}{2}

Verify Solutions:

x = \frac{1 + 41}{2}

True

, x = \frac{1 - 41}{2}

False

The solution is

x = \frac{1 + 41}{2}

ln (x) = 900

lo g_{10} (x^{2} + 36) = 2

ln (x) = 5.014055365

0.8 = ln (x)

lo g_{7} (10 - 6 x) = 4

What is the answer to log_{2}(x+1)+log_{2}(x-2)=3 ?
The answer to log_{2}(x+1)+log_{2}(x-2)=3 is x=(1+sqrt(41))/2
Solve for x: log_{2}(x+1)+log_{2}(x-2)=3
The solution is x=(1+sqrt(41))/2