Michael J. answered • 04/22/16

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Using the properties of logs, we can condense the equations so that we have a single log. Also, the solution to a logarithm is the exponent of the log's base number.

1)

log

_{6}(x(x + 1)) = 1log

_{6}(x^{2}+ x) = 16

^{1}= x^{2}+ x6 = x

^{2}+ xSubtract 6 on both sides of the equation.

0 = x

^{2}+ x - 6Now you have a quadratic equation. Factoring, we get

0 = (x + 3)(x - 2)

Solving for x from this equation.

x = -3 and x = 2

Plug in these values into the equation to validate them.

2)

log

_{6}((x + 2)(x + 3)) = 1log

_{6}(x^{2}+ 5x + 6) = 16

^{1}= x^{2}+ 5x + 66 = x

^{2}+ 5x + 6x

^{2}+ 5x = 0We have a quadratic equation again. Factoring, we get

x(x + 5) = 0

Solve for x. I leave this part to you to finish.