## Precalculus (6th Edition) Blitzer

Solution set = $\{23\}$.
$\log_{4}(3x-5)=3$ Logarithms are defined only for positive arguments, so any solution must satisfy $3x-5 \gt 0\Rightarrow\qquad x \gt 5/3\qquad(*)$ Write the RHS as $\log_{4}($....$)$ using the basic property $\log_{b}b^{x}=x$ $\log_{4}(3x-5)=\log_{4}4^{3}$ logarithmic functions are one-to-one; if $\log_{b}M=\log_{b}N$, then M=N $3x-5=4^{3}$ $3x-5=64\qquad$... add 5 $3x=69\qquad$... divide with 3 $x=23$ which satisfies the condition (*), and is a valid solution. Solution set = $\{23\}$.