Question

please help solve this pre cal problem!

ln8-ln(x-9)=4

Answer 1

& please show steps so I can learn :)

Answer 2

\(\bf ln(8)-ln(x-9)=4\implies ln\left(\cfrac{8}{x-9}\right)=4\implies log_e\left(\cfrac{8}{x-9}\right)=4
\\ \quad \\
\textit{log cancellation rule of }\large a^{log_ax}=x\qquad thus
\\ \quad \\

\large log_e\left(\cfrac{8}{x-9}\right)=4\implies {\color{red}{ e}}^{log_e\left(\frac{8}{x-9}\right)}={\color{red}{ e}}^4\implies

\cfrac{8}{x-9}=e^4\)

Answer 3

surely you can take it from there

Answer 4

from there would i multiply both sides by x-9?

Answer 5

yeap, cross-multiply and solve for "x"

Answer 6

thank you so much!! :)

Answer 7

yw