Question

Help:
Solve equation and find the exact solution
ln(x-3)=ln(2x-9)

Answer 1

\(\bf ln(x-3)=ln(2x-9)\\ \quad \\
\textit{let's take natural logarithm to both sides}\quad :)\\
x-3 = 2x-9\)

Answer 2

well... ahemm.... anyhow.. not take it ... just raised lemme give you the longer version anyway hehe

Answer 3

\(\bf ln(x-3)=ln(2x-9)\\ \quad \\
\textit{exponentialize by "e"}\\
e^{ln(x-3)}=e^{ln(2x-9)}\\ \quad \\
\textit{log cancellation rule of }\quad a^{log_ax} = x\\ \quad \\
e^{ln(x-3)}=e^{ln(2x-9)}\implies e^{log_e(x-3)}=e^{log_e(2x-9)}\\ \quad \\
x-3 = 2x-9\)

Answer 4

Thank you so much!(:

Answer 5

yw

Answer 6

Can I ask you one more thing.
what if it is like:
\[\log _{2}(\log _{3}(\log _{4}(x)))=0\]

Answer 7

well, we know that a logarithmic notation is just another way to write an exponential notation

so say \(\bf log _{2}(log _{3}(log _{4}(x)))=0\\ \quad \\ \quad \\
log_2(\square) = 0\implies 2^0 =\square\\
\textit{well, we know that }\quad 2^0 = 1\quad thus\quad \square =1\\ \quad \\
log _{3}(log _{4}(x))=1\\ \quad \\
log_3(\square) = 1\implies 3^1 = \square\\
\textit{well, wel know that }\quad 3^1 =3\quad thus \quad \square=3\\ \quad \\
log _{4}(x) = 3 \)

Answer 8

and then we do like we did above, we exponentialize "4", so that the Coefficient and the Base of the Log match

that is, we have log base 4, we use a "4" to match it up and use the log cancellation rule

Answer 9

\(\bf \large {log _{4}(x) = 3\\
\textit{log cancellation rule of }a^{log_ax} =x\\ \quad \\
log _4(x) = 3\implies 4^{log _4(x) }=4^3}\)

can you tell what's "x"?