Solve the equation and express the solution in exact form.
-$\ln 9 x + \ln 7 x = \ln 64$
A) $\left\{ \frac { e ^ { 64 } } { 63 } \right\}$
B) $\{ 1 \}$
C) $\{ 0 \}$
D) $\left\{ \left( \frac { 64 } { 63 } \right) ^ { 1 / 2 } \right\}$

Question

Quizplus · Accepted Answer

Using the properties of logarithms, we can combine the left side of the equation: $\ln(9x) + \ln(7x) = \ln(63x^2)$. This is set equal to $\ln(64)$, so we have $\ln(63x^2) = \ln(64)$. Setting the arguments of the logarithms equal to each other gives $63x^2 = 64$. Solving for $x$ gives $x^2 = \frac{64}{63}$, so $x = \pm\sqrt{\frac{64}{63}}$. However, since $x$ is in the argument of a logarithm, it must be positive, leaving $x = \sqrt{\frac{64}{63}}$.

Solve the Equation and Express the Solution in Exact Form ln⁡9x+ln⁡7x=ln⁡64\ln 9 x + \ln 7 x = \ln 64ln9x+ln7x=ln64

Solve the Equation and Express the Solution in Exact Form $\ln 9 x + \ln 7 x = \ln 64$