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`log_2x+log_2 (x+1) = 1`
To solve for x, apply the product rule of logarithm which is `log_am + log_an = log_a(m*n)` .
`log_2 (x(x+1)) = 1`
Then, express the equation to its equivalent exponential form.
Note that `log_a b=m` can be written as `a^m =b` .
`2^1 = x^2+x`
Express the equation in quadratic form `ax^2+bx+c=0` .
Set each factor to zero and solve for x.
`x+2=0` and `x-1=0`
To check, substitute the values of x to the original equation.
`x=-2` , `log_2 (-2)+log_2(-2+1)=1`
`log_2 -2+log_2-1=1` (Invalid, because negative argument of logarithm is not allowed.)
x=1, `log_2 1+log_2(1+1) = 1`
`log_2 1+log_2 = 1`
`log_2 2 = 1`
`1 = 1` (True)
Hence, the solution to the equation `log_2x + log_2(x+1)= 1` is `x=1` .
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