Answer
2
Work Step by Step
The given equation:
$\log_6x+\log_6(x+1)=1$
The equation is rewritten using a logarithm rule:
$\log_6(x(x+1))=1$
The logarithm is expanded:
$\log_6(x^2 + x)=1$
1 is rewritten as a logarithm:
$\log_6(x^2 + x)=\log_66$
By comparing the "insides" of the logarithms, the equation is simplified to:
$x^2 + x = 6$
Rearrange:
$x^2 + x - 6 =0 $
Solve quadratic using technology, quadratic formula, or factorization to find:
$x=2$
$x=-3$
Reject the second x-value because x cannot be negative, being inside a logarithm.
Thus $x=2$.
