Solve: $\log _ { 6 } x + \log _ { 6 } ( x - 20 ) = \log _ { 6 } 189$
A) {27}
B) {7}
C) $\left\{ \frac { 209 } { 2 } \right\}$
D) no solution

Question

Quizplus · Accepted Answer

Using the property of logarithms that $\log_a b + \log_a c = \log_a (bc)$, we can combine the left side of the equation: $\log_6 x + \log_6 (x - 20) = \log_6 (x(x - 20))$. This simplifies to $\log_6 (x^2 - 20x) = \log_6 189$. Since the bases of the logarithms are the same, we can equate the arguments: $x^2 - 20x = 189$. Rearranging gives $x^2 - 20x - 189 = 0$. Factoring or using the quadratic formula, we find $x = 27$ or $x = -7$. However, $x = -7$ is not a valid solution because it would result in a negative argument for the logarithm, which is undefined. Therefore, the only solution is $x = 27$.

Solve log⁡6x+log⁡6(x−20)=log⁡6189\log _ { 6 } x + \log _ { 6 } ( x - 20 ) = \log _ { 6 } 189log6​x+log6​(x−20)=log6​189

Solve $\log _ { 6 } x + \log _ { 6 } ( x - 20 ) = \log _ { 6 } 189$