What is the answer to 2+log_{2}(2x+2)=0 ?

The answer to 2+log_{2}(2x+2)=0 is x=-7/8

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2+log_{2}(2x+2)=0

2 + lo g_{2} (2 x + 2) = 0

x = - \frac{7}{8}

Solution steps

2 + lo g_{2} (2 x + 2) = 0

Move

2

to the right side

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

Apply log rules

2 x + 2 = \frac{1}{4}

Solve

2 x + 2 = \frac{1}{4} : x = - \frac{7}{8}

x = - \frac{7}{8}

Verify Solutions:

x = - \frac{7}{8}

True

The solution is

x = - \frac{7}{8}

x, y = ln (x^{2})

0.789 = ln (x^{3})

p \cdot ln (A) = ln (A^{p})

3 ln (2 x) = 6

x, 5 + lo g_{2} (- 5 x) = 9

What is the answer to 2+log_{2}(2x+2)=0 ?
The answer to 2+log_{2}(2x+2)=0 is x=-7/8
Solve for x: 2+log_{2}(2x+2)=0
The solution is x=-7/8