What is the answer to log_{10}(2)(x+2)+log_{10}(2)(x-1)=2 ?

The answer to log_{10}(2)(x+2)+log_{10}(2)(x-1)=2 is x=(2-log_{10}(2))/(2log_{10)(2)}

Solve for x: log_{10}(2)(x+2)+log_{10}(2)(x-1)=2

The solution is x=(2-log_{10}(2))/(2log_{10)(2)}

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lo g_{10} (2) (x + 2) + lo g_{10} (2) (x - 1) = 2

x = \frac{2 - lo g _{10} ( 2 )}{2 lo g _{10} ( 2 )}

Decimal Notation

x = 2.82192 \dots

Solution steps

lo g_{10} (2) (x + 2) + lo g_{10} (2) (x - 1) = 2

Expand

lo g_{10} (2) (x + 2) + lo g_{10} (2) (x - 1) : 2 lo g_{10} (2) x + lo g_{10} (2)

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

Move

lo g_{10} (2)

to the right side

2 lo g_{10} (2) x = 2 - lo g_{10} (2)

Divide both sides by

2 lo g_{10} (2)

x = \frac{2 - lo g _{10} ( 2 )}{2 lo g _{10} ( 2 )}

- \frac{1}{5} x - 25 = - 9

2 x - 2 = 6

d, 2 m^{3} d m + 3 m n^{3} d n = n^{3} d m

\frac{x + 2}{6} = \frac{x - 4}{7}

f = 3 (2)^{- 1}

What is the answer to log_{10}(2)(x+2)+log_{10}(2)(x-1)=2 ?
The answer to log_{10}(2)(x+2)+log_{10}(2)(x-1)=2 is x=(2-log_{10}(2))/(2log_{10)(2)}
Solve for x: log_{10}(2)(x+2)+log_{10}(2)(x-1)=2
The solution is x=(2-log_{10}(2))/(2log_{10)(2)}